Chapter P: Fundamental Concepts of Algebra

Chapter P Prerequisites Fundamental Concepts of Algebra

What can algebra possibly have to tell me about

  • the skyrocketing cost of a college education?

  • student-loan debt?

  • my workouts?

  • the effects of alcohol?

  • the meaning of the national debt that is more than $25 trillion?

  • time dilation on a futuristic high-speed journey to a nearby star?

  • ethnic diversity in the United States?

  • the widening imbalance between numbers of women and men on college campuses?

This chapter reviews fundamental concepts of algebra that are prerequisites for the study of college algebra. Throughout the chapter, you will see how the special language of algebra describes your world.

Here’s where you’ll find these applications:

Section P.1: Algebraic Expressions, Mathematical Models, and Real Numbers

Section P.1 Algebraic Expressions, Mathematical Models, and Real Numbers

Learning Objectives

What You’ll Learn

  1. 1 Evaluate algebraic expressions.

  2. 2 Use mathematical models.

  3. 3 Find the intersection of two sets.

  4. 4 Find the union of two sets.

  5. 5 Recognize subsets of the real numbers.

  6. 6 Use inequality symbols.

  7. 7 Evaluate absolute value.

  8. 8 Use absolute value to express distance.

  9. 9 Identify properties of the real numbers.

  10. 10 Simplify algebraic expressions.

How would your lifestyle change if a gallon of gas cost $9.15? Or if the price of a staple such as milk was $15? That’s how much those products would cost if their prices had increased at the same rate college tuition has increased since 1980. (Source: Center for College Affordability and Productivity) In this section, you will learn how the special language of algebra describes your world, including the skyrocketing cost of a college education.

Algebraic Expressions

Algebra uses letters, such as x and y, to represent numbers. If a letter is used to represent various numbers, it is called a variable. For example, imagine that you are basking in the sun on the beach. We can let x represent the number of minutes that you can stay in the sun without burning with no sunscreen. With a number 6 sunscreen, exposure time without burning is six times as long, or 6 times x. This can be written 6x, but it is usually expressed as 6x. Placing a number and a letter next to one another indicates multiplication.

Notice that 6x combines the number 6 and the variable x using the operation of multiplication. A combination of variables and numbers using the operations of addition, subtraction, multiplication, or division, as well as powers or roots, is called an algebraic expression. Here are some examples of algebraic expressions:

x+6,x6,6x,x6,3x+5,x23,x+7.

Many algebraic expressions involve exponents. For example, the algebraic expression

x2+361x+3193

approximates the average cost of tuition and fees at public U.S. colleges for the school year ending x years after 2000. The expression x2 means xx and is read “x to the second power” or “x squared.” The exponent, 2, indicates that the base, x, appears as a factor two times. The negative sign in front of x2 indicates that x2 is multiplied by 1.

Exponential Notation

If n is a counting number (1, 2, 3, and so on),

B to the n t h power, where b is labeled, base and n is labeled, exponent or power, = b times b times b times ellipsis times b. The right side of the equation is labeled, b appears as a factor n times.

bn is read “the nth power of b” or “b to the nth power.” Thus, the nth power of b is defined as the product of n factors of b. The expression bn  is called an exponential expression. Furthermore, b1=b.

For example,

82=88=64,53=555=125,and24=2222=16.
Objective 1: Evaluate algebraic expressions

Evaluating Algebraic Expressions

  1. Objective 1Evaluate algebraic expressions.

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Evaluating an algebraic expression means to find the value of the expression for a given value of the variable.

Many algebraic expressions involve more than one operation. Evaluating an algebraic expression without a calculator involves carefully applying the following order of operations agreement:

The Order of Operations Agreement

  1. Perform operations within the innermost parentheses and work outward. If the algebraic expression involves a fraction, treat the numerator and the denominator as if they were each enclosed in parentheses.

  2. Evaluate all exponential expressions.

  3. Perform multiplications and divisions as they occur, working from left to right.

  4. Perform additions and subtractions as they occur, working from left to right.

Example 1 Evaluating an Algebraic Expression

Evaluate 7+5(x4)3 for x=6.

Solution

7+5(x4)3=7+5(64)3Replace x with 6.=7+5(2)3First work inside parentheses:  64=2.=7+5(8)Evaluate the exponential expression: 23=222=8.=7+40Multiply: 5(8)=40.=47Add.

Check Point 1

  • Evaluate 8+6(x3)2 for x=13.

Objective 2: Use mathematical models

Formulas and Mathematical Models

  1. Objective 2Use mathematical models.

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An equation is formed when an equal sign is placed between two algebraic expressions. One aim of algebra is to provide a compact, symbolic description of the world. These descriptions involve the use of formulas. A formula is an equation that uses variables to express a relationship between two or more quantities.

Here are two examples of formulas related to heart rate and exercise.

Two photos. The first photo depicts a woman seated on the floor with coke and popcorn. The second photo depicts three women and a man, kicking bags in a gym.
P.1-3 Full Alternative Text

The process of finding formulas to describe real-world phenomena is called mathematical modeling. Such formulas, together with the meaning assigned to the variables, are called mathematical models. We often say that these formulas model, or describe, the relationships among the variables.

Example 2 Modeling the Cost of Attending a Public College

The bar graph in Figure P.1 shows the average cost of tuition and fees for public four-year colleges, adjusted for inflation. The formula

T=x2+361x+3193

models the average cost of tuition and fees, T, for public U.S. colleges for the school year ending x years after 2000.

  1. Use the formula to find the average cost of tuition and fees at public U.S. colleges for the school year ending in 2020.

  2. By how much does the formula underestimate or overestimate the actual cost shown in Figure P.1?

Figure P.1

A young boy with a big size calculator and a bar graph titled, average cost of tuition and fees at public four year U S colleges.

Source: The College Board

Solution

  1. Because 2020 is 20 years after 2000, we substitute 20 for x in the given formula. Then we use the order of operations to find T, the average cost of tuition and fees for the school year ending in 2020.

    Solution to simplify T =negative x squared + 361 times + 3193.

    The formula indicates that for the school year ending in 2020, the average cost of tuition and fees at public U.S. colleges was $10,013.

  2. Figure P.1 shows that the average cost of tuition and fees for the school year ending in 2020 was $10,440.

    The cost obtained from the formula, $10,013, underestimates the actual data value by $10,440$10,013, or by $427.

Check Point 2

  1. Use the formula T=x2+361x+3193, described in Example 2, to find the average cost of tuition and fees at public U.S. colleges for the school year ending in 2016.

  2. By how much does the formula underestimate or overestimate the actual cost shown in Figure P.1?

Sometimes a mathematical model gives an estimate that is not a good approximation or is extended to include values of the variable that do not make sense. In these cases, we say that model breakdown has occurred. For example, it is not likely that the formula in Example 2 would give a good estimate of tuition and fees in 2050 because it is too far in the future. Thus, model breakdown would occur.

Objective 2: Use mathematical models

Formulas and Mathematical Models

  1. Objective 2Use mathematical models.

Watch Video

An equation is formed when an equal sign is placed between two algebraic expressions. One aim of algebra is to provide a compact, symbolic description of the world. These descriptions involve the use of formulas. A formula is an equation that uses variables to express a relationship between two or more quantities.

Here are two examples of formulas related to heart rate and exercise.

Two photos. The first photo depicts a woman seated on the floor with coke and popcorn. The second photo depicts three women and a man, kicking bags in a gym.
P.1-3 Full Alternative Text

The process of finding formulas to describe real-world phenomena is called mathematical modeling. Such formulas, together with the meaning assigned to the variables, are called mathematical models. We often say that these formulas model, or describe, the relationships among the variables.

Example 2 Modeling the Cost of Attending a Public College

The bar graph in Figure P.1 shows the average cost of tuition and fees for public four-year colleges, adjusted for inflation. The formula

T=x2+361x+3193

models the average cost of tuition and fees, T, for public U.S. colleges for the school year ending x years after 2000.

  1. Use the formula to find the average cost of tuition and fees at public U.S. colleges for the school year ending in 2020.

  2. By how much does the formula underestimate or overestimate the actual cost shown in Figure P.1?

Figure P.1

A young boy with a big size calculator and a bar graph titled, average cost of tuition and fees at public four year U S colleges.

Source: The College Board

Solution

  1. Because 2020 is 20 years after 2000, we substitute 20 for x in the given formula. Then we use the order of operations to find T, the average cost of tuition and fees for the school year ending in 2020.

    Solution to simplify T =negative x squared + 361 times + 3193.

    The formula indicates that for the school year ending in 2020, the average cost of tuition and fees at public U.S. colleges was $10,013.

  2. Figure P.1 shows that the average cost of tuition and fees for the school year ending in 2020 was $10,440.

    The cost obtained from the formula, $10,013, underestimates the actual data value by $10,440$10,013, or by $427.

Check Point 2

  1. Use the formula T=x2+361x+3193, described in Example 2, to find the average cost of tuition and fees at public U.S. colleges for the school year ending in 2016.

  2. By how much does the formula underestimate or overestimate the actual cost shown in Figure P.1?

Sometimes a mathematical model gives an estimate that is not a good approximation or is extended to include values of the variable that do not make sense. In these cases, we say that model breakdown has occurred. For example, it is not likely that the formula in Example 2 would give a good estimate of tuition and fees in 2050 because it is too far in the future. Thus, model breakdown would occur.

Objective 3: Find the intersection of two sets

Sets

  1. Objective 3Find the intersection of two sets.

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Before we describe the set of real numbers, let’s be sure you are familiar with some basic ideas about sets. A set is a collection of objects whose contents can be clearly determined. The objects in a set are called the elements of the set. For example, the set of numbers used for counting can be represented by

{1, 2, 3, 4, 5,}.

The braces, { }, indicate that we are representing a set. This form of representation, called the roster method, uses commas to separate the elements of the set. The symbol consisting of three dots after the 5, called an ellipsis, indicates that there is no final element and that the listing goes on forever.

A set can also be written in set-builder notation. In this notation, the elements of the set are described but not listed. Here is an example:

Left brace x vertical bar x is a counting number less than 6 right brace. x is labeled, the set of all x. The vertical bar is labeled, such that. x is a counting number less than 6 is labeled, x is a counting number less than 6.

The same set written using the roster method is

{1, 2, 3, 4, 5}.

If A and B are sets, we can form a new set consisting of all elements that are in both A and B. This set is called the intersection of the two sets.

Definition of the Intersection of Sets

The intersection of sets A and B, written AB, is the set of elements common to both set A and set B. This definition can be expressed in set-builder notation as follows:

AB={x|x is an element of A AND x is an element of B}.

Figure P.2 shows a useful way of picturing the intersection of sets A and B. The figure indicates that AB contains those elements that belong to both A and B at the same time.

Figure P.2 Picturing the intersection of two sets

Two intersecting circles, A and B, with their areas shaded and the intersecting region labeled, the intersection of A and B.

Example 3 Finding the Intersection of Two Sets

Find the intersection: {7, 8, 9, 10, 11}{6, 8, 10, 12}.

Solution

The elements common to {7, 8, 9, 10, 11} and {6, 8, 10, 12} are 8 and 10. Thus,

{7, 8, 9, 10, 11}{6, 8, 10, 12}={8, 10}.

Check Point 3

  • Find the intersection: {3, 4, 5, 6, 7}{3, 7, 8, 9}.

If a set has no elements, it is called the empty set, or the null set, and is represented by the symbol Ø. Here is an example that shows how the empty set can result when finding the intersection of two sets:

The intersection of the set of 2, 4, 6 and the set of 3, 5, 7 = null set. A text reads, These sets have no elements in common. Their intersection has no elements and is the empty set.
Objective 5: Recognize subsets of the real numbers

The Set of Real Numbers

  1. Objective 5Recognize subsets of the real numbers.

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The sets that make up the real numbers are summarized in Table P.1. We refer to these sets as subsets of the real numbers, meaning that all elements in each subset are also elements in the set of real numbers.

Table P.1 Important Subsets of the Real Numbers

A table titled, Important Subsets of the Real Numbers.
Table P.1 Full Alternative Text

Notice the use of the symbol ≈ in the examples of irrational numbers. The symbol means “is approximately equal to.” Thus,

21.414214.

We can verify that this is only an approximation by multiplying 1.414214 by itself. The product is very close to, but not exactly, 2:

1.414214×1.414214=2.000001237796.

Not all square roots are irrational. For example, 25=5 because 52=55=25. Thus, 25  is a natural number, a whole number, an integer, and a rational number (25=51).

The set of real numbers is formed by taking the union of the sets of rational numbers and irrational numbers. Thus, every real number is either rational or irrational, as shown in Figure P.4.

Figure P.4 Every real number is either rational or irrational.

A horizontal rectangle labeled, real numbers are split into two vertical rectangles, rational numbers, and irrational numbers.
Figure P.4 Full Alternative Text

Real Numbers

The set of real numbers is the set of numbers that are either rational or irrational:

{x|x is rational or x is irrational}.

The symbol ℝ is used to represent the set of real numbers. Thus,

={x|x is rational}{x|x is irrational}.

Example 5 Recognizing Subsets of the Real Numbers

Consider the following set of numbers:

{7, 34, 0, 0.6¯, 5, π, 7.3, 81}.

List the numbers in the set that are

  1. natural numbers.

  2. whole numbers.

  3. integers.

  4. rational numbers.

  5. irrational numbers.

  6. real numbers.

Solution

  1. Natural numbers: The natural numbers are the numbers used for counting. The only natural number in the set {7, 34 , 0, 0.6¯, 5, π, 7.3, 81} is 81 because 81=9. (9 multiplied by itself, or 92, is 81.)

  2. Whole numbers: The whole numbers consist of the natural numbers and 0. The elements of the set {7, 34, 0, 0.6¯, 5, π, 7.3, 81} that are whole numbers are 0 and 81.

  3. Integers: The integers consist of the natural numbers, 0, and the negatives of the natural numbers. The elements of the set {7, 34, 0, 0.6¯, 5, π, 7.3, 81} that are integers are 81,0, and 7.

  4. Rational numbers: All numbers in the set {7, 34, 0, 0.6¯, 5, π, 7.3, 81} that can be expressed as the quotient of integers are rational numbers. These include 7(7=71), 34, 0(0=01), and 81(81=91). Furthermore, all numbers in the set that are terminating or repeating decimals are also rational numbers. These include 0.6¯ and 7.3.

  5. Irrational numbers: The irrational numbers in the set {7, 34, 0, 0.6¯, 5, π, 7.3, 81} are 5(5  2.236) and π(π  3.14). Both 5  and π are only approximately equal to 2.236 and 3.14, respectively. In decimal form, 5  and π neither terminate nor have blocks of repeating digits.

  6. Real numbers: All the numbers in the given set {7, 34, 0, 0.6¯, 5, π, 7.3, 81} are real numbers.

Check Point 5

  • Consider the following set of numbers:

    {9, 1.3, 0, 0.3¯, π2, 9 , 10}.

    List the numbers in the set that are

    1. natural numbers.

    2. whole numbers.

    3. integers.

    4. rational numbers.

    5. irrational numbers.

    6. real numbers.

The Real Number Line

The real number line is a graph used to represent the set of real numbers. An arbitrary point, called the origin, is labeled 0. Select a point to the right of 0 and label it 1. The distance from 0 to 1 is called the unit distance. Numbers to the right of the origin are positive and numbers to the left of the origin are negative. The real number line is shown in Figure P.5.

Figure P.5 The real number line

A real number line with numbers and direction to the right of 0 numbered as 1, 2, 3, 4, 5, 6, 7 labeled, positive numbers and positive direction.
Figure P.5 Full Alternative Text

Real numbers are graphed on a number line by placing a dot at the correct location for each number. The integers are easiest to locate. In Figure P.6, we’ve graphed six rational numbers and three irrational numbers on a real number line.

Figure P.6 Graphing numbers on a real number line

A real number line numbered from left to right negative 2, negative 1, 0, 1, 2, 3, 4, 5 has six rational numbers and three irrational numbers marked on it.
Figure P.6 Full Alternative Text

Every real number corresponds to a point on the number line and every point on the number line corresponds to a real number. We say that there is a one-to-one correspondence between all the real numbers and all points on a real number line.

Objective 5: Recognize subsets of the real numbers

The Set of Real Numbers

  1. Objective 5Recognize subsets of the real numbers.

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The sets that make up the real numbers are summarized in Table P.1. We refer to these sets as subsets of the real numbers, meaning that all elements in each subset are also elements in the set of real numbers.

Table P.1 Important Subsets of the Real Numbers

A table titled, Important Subsets of the Real Numbers.
Table P.1 Full Alternative Text

Notice the use of the symbol ≈ in the examples of irrational numbers. The symbol means “is approximately equal to.” Thus,

21.414214.

We can verify that this is only an approximation by multiplying 1.414214 by itself. The product is very close to, but not exactly, 2:

1.414214×1.414214=2.000001237796.

Not all square roots are irrational. For example, 25=5 because 52=55=25. Thus, 25  is a natural number, a whole number, an integer, and a rational number (25=51).

The set of real numbers is formed by taking the union of the sets of rational numbers and irrational numbers. Thus, every real number is either rational or irrational, as shown in Figure P.4.

Figure P.4 Every real number is either rational or irrational.

A horizontal rectangle labeled, real numbers are split into two vertical rectangles, rational numbers, and irrational numbers.
Figure P.4 Full Alternative Text

Real Numbers

The set of real numbers is the set of numbers that are either rational or irrational:

{x|x is rational or x is irrational}.

The symbol ℝ is used to represent the set of real numbers. Thus,

={x|x is rational}{x|x is irrational}.

Example 5 Recognizing Subsets of the Real Numbers

Consider the following set of numbers:

{7, 34, 0, 0.6¯, 5, π, 7.3, 81}.

List the numbers in the set that are

  1. natural numbers.

  2. whole numbers.

  3. integers.

  4. rational numbers.

  5. irrational numbers.

  6. real numbers.

Solution

  1. Natural numbers: The natural numbers are the numbers used for counting. The only natural number in the set {7, 34 , 0, 0.6¯, 5, π, 7.3, 81} is 81 because 81=9. (9 multiplied by itself, or 92, is 81.)

  2. Whole numbers: The whole numbers consist of the natural numbers and 0. The elements of the set {7, 34, 0, 0.6¯, 5, π, 7.3, 81} that are whole numbers are 0 and 81.

  3. Integers: The integers consist of the natural numbers, 0, and the negatives of the natural numbers. The elements of the set {7, 34, 0, 0.6¯, 5, π, 7.3, 81} that are integers are 81,0, and 7.

  4. Rational numbers: All numbers in the set {7, 34, 0, 0.6¯, 5, π, 7.3, 81} that can be expressed as the quotient of integers are rational numbers. These include 7(7=71), 34, 0(0=01), and 81(81=91). Furthermore, all numbers in the set that are terminating or repeating decimals are also rational numbers. These include 0.6¯ and 7.3.

  5. Irrational numbers: The irrational numbers in the set {7, 34, 0, 0.6¯, 5, π, 7.3, 81} are 5(5  2.236) and π(π  3.14). Both 5  and π are only approximately equal to 2.236 and 3.14, respectively. In decimal form, 5  and π neither terminate nor have blocks of repeating digits.

  6. Real numbers: All the numbers in the given set {7, 34, 0, 0.6¯, 5, π, 7.3, 81} are real numbers.

Check Point 5

  • Consider the following set of numbers:

    {9, 1.3, 0, 0.3¯, π2, 9 , 10}.

    List the numbers in the set that are

    1. natural numbers.

    2. whole numbers.

    3. integers.

    4. rational numbers.

    5. irrational numbers.

    6. real numbers.

The Real Number Line

The real number line is a graph used to represent the set of real numbers. An arbitrary point, called the origin, is labeled 0. Select a point to the right of 0 and label it 1. The distance from 0 to 1 is called the unit distance. Numbers to the right of the origin are positive and numbers to the left of the origin are negative. The real number line is shown in Figure P.5.

Figure P.5 The real number line

A real number line with numbers and direction to the right of 0 numbered as 1, 2, 3, 4, 5, 6, 7 labeled, positive numbers and positive direction.
Figure P.5 Full Alternative Text

Real numbers are graphed on a number line by placing a dot at the correct location for each number. The integers are easiest to locate. In Figure P.6, we’ve graphed six rational numbers and three irrational numbers on a real number line.

Figure P.6 Graphing numbers on a real number line

A real number line numbered from left to right negative 2, negative 1, 0, 1, 2, 3, 4, 5 has six rational numbers and three irrational numbers marked on it.
Figure P.6 Full Alternative Text

Every real number corresponds to a point on the number line and every point on the number line corresponds to a real number. We say that there is a one-to-one correspondence between all the real numbers and all points on a real number line.

Objective 5: Recognize subsets of the real numbers

The Set of Real Numbers

  1. Objective 5Recognize subsets of the real numbers.

Watch Video

The sets that make up the real numbers are summarized in Table P.1. We refer to these sets as subsets of the real numbers, meaning that all elements in each subset are also elements in the set of real numbers.

Table P.1 Important Subsets of the Real Numbers

A table titled, Important Subsets of the Real Numbers.
Table P.1 Full Alternative Text

Notice the use of the symbol ≈ in the examples of irrational numbers. The symbol means “is approximately equal to.” Thus,

21.414214.

We can verify that this is only an approximation by multiplying 1.414214 by itself. The product is very close to, but not exactly, 2:

1.414214×1.414214=2.000001237796.

Not all square roots are irrational. For example, 25=5 because 52=55=25. Thus, 25  is a natural number, a whole number, an integer, and a rational number (25=51).

The set of real numbers is formed by taking the union of the sets of rational numbers and irrational numbers. Thus, every real number is either rational or irrational, as shown in Figure P.4.

Figure P.4 Every real number is either rational or irrational.

A horizontal rectangle labeled, real numbers are split into two vertical rectangles, rational numbers, and irrational numbers.
Figure P.4 Full Alternative Text

Real Numbers

The set of real numbers is the set of numbers that are either rational or irrational:

{x|x is rational or x is irrational}.

The symbol ℝ is used to represent the set of real numbers. Thus,

={x|x is rational}{x|x is irrational}.

Example 5 Recognizing Subsets of the Real Numbers

Consider the following set of numbers:

{7, 34, 0, 0.6¯, 5, π, 7.3, 81}.

List the numbers in the set that are

  1. natural numbers.

  2. whole numbers.

  3. integers.

  4. rational numbers.

  5. irrational numbers.

  6. real numbers.

Solution

  1. Natural numbers: The natural numbers are the numbers used for counting. The only natural number in the set {7, 34 , 0, 0.6¯, 5, π, 7.3, 81} is 81 because 81=9. (9 multiplied by itself, or 92, is 81.)

  2. Whole numbers: The whole numbers consist of the natural numbers and 0. The elements of the set {7, 34, 0, 0.6¯, 5, π, 7.3, 81} that are whole numbers are 0 and 81.

  3. Integers: The integers consist of the natural numbers, 0, and the negatives of the natural numbers. The elements of the set {7, 34, 0, 0.6¯, 5, π, 7.3, 81} that are integers are 81,0, and 7.

  4. Rational numbers: All numbers in the set {7, 34, 0, 0.6¯, 5, π, 7.3, 81} that can be expressed as the quotient of integers are rational numbers. These include 7(7=71), 34, 0(0=01), and 81(81=91). Furthermore, all numbers in the set that are terminating or repeating decimals are also rational numbers. These include 0.6¯ and 7.3.

  5. Irrational numbers: The irrational numbers in the set {7, 34, 0, 0.6¯, 5, π, 7.3, 81} are 5(5  2.236) and π(π  3.14). Both 5  and π are only approximately equal to 2.236 and 3.14, respectively. In decimal form, 5  and π neither terminate nor have blocks of repeating digits.

  6. Real numbers: All the numbers in the given set {7, 34, 0, 0.6¯, 5, π, 7.3, 81} are real numbers.

Check Point 5

  • Consider the following set of numbers:

    {9, 1.3, 0, 0.3¯, π2, 9 , 10}.

    List the numbers in the set that are

    1. natural numbers.

    2. whole numbers.

    3. integers.

    4. rational numbers.

    5. irrational numbers.

    6. real numbers.

The Real Number Line

The real number line is a graph used to represent the set of real numbers. An arbitrary point, called the origin, is labeled 0. Select a point to the right of 0 and label it 1. The distance from 0 to 1 is called the unit distance. Numbers to the right of the origin are positive and numbers to the left of the origin are negative. The real number line is shown in Figure P.5.

Figure P.5 The real number line

A real number line with numbers and direction to the right of 0 numbered as 1, 2, 3, 4, 5, 6, 7 labeled, positive numbers and positive direction.
Figure P.5 Full Alternative Text

Real numbers are graphed on a number line by placing a dot at the correct location for each number. The integers are easiest to locate. In Figure P.6, we’ve graphed six rational numbers and three irrational numbers on a real number line.

Figure P.6 Graphing numbers on a real number line

A real number line numbered from left to right negative 2, negative 1, 0, 1, 2, 3, 4, 5 has six rational numbers and three irrational numbers marked on it.
Figure P.6 Full Alternative Text

Every real number corresponds to a point on the number line and every point on the number line corresponds to a real number. We say that there is a one-to-one correspondence between all the real numbers and all points on a real number line.

Objective 7: Evaluate absolute value

Absolute Value

  1. Objective 7Evaluate absolute value.

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The absolute value of a real number a, denoted by |a|, is the distance from 0 to a on the number line. This distance is always taken to be nonnegative. For example, the real number line in Figure P.8 shows that

|3|=3and|5|=5.

Figure P.8 Absolute value as the distance from 0

A number line numbered from left to right as negative 5, negative 4, negative 3, negative 2, negative 1, 0, 1, 2, 3, 4, 5.
Figure P.8 Full Alternative Text

The absolute value of 3 is 3 because 3 is 3 units from 0 on the number line. The absolute value of 5 is 5 because 5 is 5 units from 0 on the number line. The absolute value of a positive real number or 0 is the number itself. The absolute value of a negative real number, such as 3, is the number without the negative sign.

We can define the absolute value of the real number x without referring to a number line. The algebraic definition of the absolute value of x is given as follows:

Definition of Absolute Value

|x|={xifx0 xifx<0 

If x is nonnegative (that is, x0), the absolute value of x is the number itself. For example,

The image shows examples of absolute values.
P.1-15 Full Alternative Text

If x is a negative number (that is, x<0), the absolute value of x is the opposite of x. This makes the absolute value positive. For example,

Examples of absolute values.
P.1-16 Full Alternative Text

Observe that the absolute value of any nonzero number is always positive.

Example 6 Evaluating Absolute Value

Rewrite each expression without absolute value bars:

  1. |31|

  2. |2π|

  3. | x |x if x<0.

Solution

  1. Because 3 1.7, the number inside the absolute value bars, 31, is positive. The absolute value of a positive number is the number itself. Thus,

    |31|=31.
  2. Because π3.14, the number inside the absolute value bars, 2π, is negative. The absolute value of x when x<0 is x. Thus,

    |2π|=(2π)=π2.
  3. If x<0, then |x|=x. Thus,

    |x| x=xx=1.

Check Point 6

  • Rewrite each expression without absolute value bars:

    1. |12|

    2. |π3|

    3. |x|x if x>0.

Listed below are several basic properties of absolute value. Each of these properties can be derived from the definition of absolute value.

Properties of Absolute Value

For all real numbers a and b,

  1. |a|0

  2. |a|=|a|

  3. a|a|

  4. |ab|=|a||b|

  5. |a b|=|a| |b|, b0

  6. |a+b||a|+|b| (called the triangle inequality).

Objective 7: Evaluate absolute value

Absolute Value

  1. Objective 7Evaluate absolute value.

Watch Video

The absolute value of a real number a, denoted by |a|, is the distance from 0 to a on the number line. This distance is always taken to be nonnegative. For example, the real number line in Figure P.8 shows that

|3|=3and|5|=5.

Figure P.8 Absolute value as the distance from 0

A number line numbered from left to right as negative 5, negative 4, negative 3, negative 2, negative 1, 0, 1, 2, 3, 4, 5.
Figure P.8 Full Alternative Text

The absolute value of 3 is 3 because 3 is 3 units from 0 on the number line. The absolute value of 5 is 5 because 5 is 5 units from 0 on the number line. The absolute value of a positive real number or 0 is the number itself. The absolute value of a negative real number, such as 3, is the number without the negative sign.

We can define the absolute value of the real number x without referring to a number line. The algebraic definition of the absolute value of x is given as follows:

Definition of Absolute Value

|x|={xifx0 xifx<0 

If x is nonnegative (that is, x0), the absolute value of x is the number itself. For example,

The image shows examples of absolute values.
P.1-15 Full Alternative Text

If x is a negative number (that is, x<0), the absolute value of x is the opposite of x. This makes the absolute value positive. For example,

Examples of absolute values.
P.1-16 Full Alternative Text

Observe that the absolute value of any nonzero number is always positive.

Example 6 Evaluating Absolute Value

Rewrite each expression without absolute value bars:

  1. |31|

  2. |2π|

  3. | x |x if x<0.

Solution

  1. Because 3 1.7, the number inside the absolute value bars, 31, is positive. The absolute value of a positive number is the number itself. Thus,

    |31|=31.
  2. Because π3.14, the number inside the absolute value bars, 2π, is negative. The absolute value of x when x<0 is x. Thus,

    |2π|=(2π)=π2.
  3. If x<0, then |x|=x. Thus,

    |x| x=xx=1.

Check Point 6

  • Rewrite each expression without absolute value bars:

    1. |12|

    2. |π3|

    3. |x|x if x>0.

Listed below are several basic properties of absolute value. Each of these properties can be derived from the definition of absolute value.

Properties of Absolute Value

For all real numbers a and b,

  1. |a|0

  2. |a|=|a|

  3. a|a|

  4. |ab|=|a||b|

  5. |a b|=|a| |b|, b0

  6. |a+b||a|+|b| (called the triangle inequality).

Objective 9: Identify properties of the real numbers

Properties of Real Numbers and Algebraic Expressions

  1. Objective 9Identify properties of the real numbers.

Watch Video

When you use your calculator to add two real numbers, you can enter them in any order. The fact that two real numbers can be added in any order is called the commutative property of addition. You probably use this property, as well as other properties of real numbers listed in Table P.2, without giving it much thought. The properties of the real numbers are especially useful when working with algebraic expressions. For each property listed in Table P.2, a, b, and c represent real numbers, variables, or algebraic expressions.

Table P.2 Properties of the Real Numbers

The table depicts the property, its meaning, and a corresponding example.
Table P.2 Full Alternative Text

The properties of the real numbers in Table P.2 apply to the operations of addition and multiplication. Subtraction and division are defined in terms of addition and multiplication.

Definitions of Subtraction and Division

Let a and b represent real numbers.

Subtraction: ab=a+(b)
We call b the additive inverse or opposite of b.

Division: a÷b=a1 b, where b0
We call 1b the multiplicative inverse or reciprocal of b. The quotient of a and b, a÷b, can be written in the form ab, where a is the numerator and b is the denominator of the fraction.

Because subtraction is defined in terms of adding an inverse, the distributive property can be applied to subtraction:

a left parenthesis b minus c right parenthesis = a b minus a c. Left parenthesis b minus c right parenthesis a = b a minus c a. Arrows extend from a to b and c, in the left hand side of both the of the equation.

For example,

4 left parenthesis 2 x minus 5 right parenthesis = 4 times 2 x minus 4 times 5 = 8 x minus 20. Arrows extend from 4 to 2 x and negative 5, in the term 4 left parenthesis 2 x minus 5 right parenthesis.
Objective 9: Identify properties of the real numbers

Properties of Real Numbers and Algebraic Expressions

  1. Objective 9Identify properties of the real numbers.

Watch Video

When you use your calculator to add two real numbers, you can enter them in any order. The fact that two real numbers can be added in any order is called the commutative property of addition. You probably use this property, as well as other properties of real numbers listed in Table P.2, without giving it much thought. The properties of the real numbers are especially useful when working with algebraic expressions. For each property listed in Table P.2, a, b, and c represent real numbers, variables, or algebraic expressions.

Table P.2 Properties of the Real Numbers

The table depicts the property, its meaning, and a corresponding example.
Table P.2 Full Alternative Text

The properties of the real numbers in Table P.2 apply to the operations of addition and multiplication. Subtraction and division are defined in terms of addition and multiplication.

Definitions of Subtraction and Division

Let a and b represent real numbers.

Subtraction: ab=a+(b)
We call b the additive inverse or opposite of b.

Division: a÷b=a1 b, where b0
We call 1b the multiplicative inverse or reciprocal of b. The quotient of a and b, a÷b, can be written in the form ab, where a is the numerator and b is the denominator of the fraction.

Because subtraction is defined in terms of adding an inverse, the distributive property can be applied to subtraction:

a left parenthesis b minus c right parenthesis = a b minus a c. Left parenthesis b minus c right parenthesis a = b a minus c a. Arrows extend from a to b and c, in the left hand side of both the of the equation.

For example,

4 left parenthesis 2 x minus 5 right parenthesis = 4 times 2 x minus 4 times 5 = 8 x minus 20. Arrows extend from 4 to 2 x and negative 5, in the term 4 left parenthesis 2 x minus 5 right parenthesis.
Objective 10: Simplify algebraic expressions

Simplifying Algebraic Expressions

  1. Objective 10Simplify algebraic expressions.

Watch Video

The terms of an algebraic expression are those parts that are separated by addition. For example, consider the algebraic expression

7x9y+z3,

which can be expressed as

7x+(9y)+z+(3).

This expression contains four terms, namely, 7x, 9y, z, and 3.

The numerical part of a term is called its coefficient. In the term 7x, the 7 is the coefficient. If a term containing one or more variables is written without a coefficient, the coefficient is understood to be 1. Thus, z means 1z. If a term is a constant, its coefficient is that constant. Thus, the coefficient of the constant term 3 is 3.

7 x + left parenthesis negative 9 y right parenthesis + z + left parenthesis negative 3 right parenthesis.
P.1-22 Full Alternative Text

The parts of each term that are multiplied are called the factors of the term. The factors of the term 7x are 7 and x.

Like terms are terms that have exactly the same variable factors. For example, 3x and 7x are like terms. The distributive property in the form

ba+ca=(b+c)a

enables us to add or subtract like terms. For example,

3x+7x=(3+7)x=10x7y2y2=7y21y2=(71)y2= 6y2.

This process is called combining like terms.

An algebraic expression is simplified when parentheses have been removed and like terms have been combined.

Example 8 Simplifying an Algebraic Expression

Simplify: 6(2x2+4x)+10(4x2+3x).

Solution

Solution to simplify 6 left parenthesis 2 x squared + 4 x right parenthesis + 10 left parenthesis 4 x squared + 3 x right parenthesis. Arrows extend from 6 and 10 to the respective x squared and x terms.

Check Point 8

  • Simplify: 7(4x2+3x)+2(5x2+x).

Properties of Negatives

The distributive property can be extended to cover more than two terms within parentheses. For example,

A solution of negative 3 left parenthesis 4 x minus 2 y + 6 right parenthesis.
P.1-24 Full Alternative Text

The voice balloons illustrate that negative signs can appear side by side. They can represent the operation of subtraction or the fact that a real number is negative. Here is a list of properties of negatives and how they are applied to algebraic expressions:

Properties of Negatives

Let a and b represent real numbers, variables, or algebraic expressions.

Property Examples
1. (1)a=a (1)4xy=4xy
2. (a)=a (6y)=6y
3. (a)b=ab (7)4xy=74xy=28xy
4. a(b)=ab 5x(3y)=5x3y=15xy
5. (a+b)=ab (7x+6y)=7x6y

6. (ab)=a+b=ba

(3x7y)=3x+7y=7y3x

It is not uncommon to see algebraic expressions with parentheses preceded by a negative sign or subtraction. Properties 5 and 6 in the box, (a+b)=ab and (ab)=a+b, are related to this situation. An expression of the form (a+b) can be simplified as follows:

A polynomial equation.
P.1-25 Full Alternative Text

Do you see a fast way to obtain the simplified expression on the right in the preceding equation? If a negative sign or a subtraction symbol appears outside parentheses, drop the parentheses and change the sign of every term within the parentheses. For example,

(3x27x4)=3x2+7x+4.

Example 9 Simplifying an Algebraic Expression

Simplify: 8x+2[5(x3)].

Solution

A solution to simplify 8 x + 2 left bracket 5 minus left parenthesis x minus 3 right parenthesis right bracket.

Check Point 9

  • Simplify: 6+4[7(x2)].

Objective 10: Simplify algebraic expressions

Simplifying Algebraic Expressions

  1. Objective 10Simplify algebraic expressions.

Watch Video

The terms of an algebraic expression are those parts that are separated by addition. For example, consider the algebraic expression

7x9y+z3,

which can be expressed as

7x+(9y)+z+(3).

This expression contains four terms, namely, 7x, 9y, z, and 3.

The numerical part of a term is called its coefficient. In the term 7x, the 7 is the coefficient. If a term containing one or more variables is written without a coefficient, the coefficient is understood to be 1. Thus, z means 1z. If a term is a constant, its coefficient is that constant. Thus, the coefficient of the constant term 3 is 3.

7 x + left parenthesis negative 9 y right parenthesis + z + left parenthesis negative 3 right parenthesis.
P.1-22 Full Alternative Text

The parts of each term that are multiplied are called the factors of the term. The factors of the term 7x are 7 and x.

Like terms are terms that have exactly the same variable factors. For example, 3x and 7x are like terms. The distributive property in the form

ba+ca=(b+c)a

enables us to add or subtract like terms. For example,

3x+7x=(3+7)x=10x7y2y2=7y21y2=(71)y2= 6y2.

This process is called combining like terms.

An algebraic expression is simplified when parentheses have been removed and like terms have been combined.

Example 8 Simplifying an Algebraic Expression

Simplify: 6(2x2+4x)+10(4x2+3x).

Solution

Solution to simplify 6 left parenthesis 2 x squared + 4 x right parenthesis + 10 left parenthesis 4 x squared + 3 x right parenthesis. Arrows extend from 6 and 10 to the respective x squared and x terms.

Check Point 8

  • Simplify: 7(4x2+3x)+2(5x2+x).

Properties of Negatives

The distributive property can be extended to cover more than two terms within parentheses. For example,

A solution of negative 3 left parenthesis 4 x minus 2 y + 6 right parenthesis.
P.1-24 Full Alternative Text

The voice balloons illustrate that negative signs can appear side by side. They can represent the operation of subtraction or the fact that a real number is negative. Here is a list of properties of negatives and how they are applied to algebraic expressions:

Properties of Negatives

Let a and b represent real numbers, variables, or algebraic expressions.

Property Examples
1. (1)a=a (1)4xy=4xy
2. (a)=a (6y)=6y
3. (a)b=ab (7)4xy=74xy=28xy
4. a(b)=ab 5x(3y)=5x3y=15xy
5. (a+b)=ab (7x+6y)=7x6y

6. (ab)=a+b=ba

(3x7y)=3x+7y=7y3x

It is not uncommon to see algebraic expressions with parentheses preceded by a negative sign or subtraction. Properties 5 and 6 in the box, (a+b)=ab and (ab)=a+b, are related to this situation. An expression of the form (a+b) can be simplified as follows:

A polynomial equation.
P.1-25 Full Alternative Text

Do you see a fast way to obtain the simplified expression on the right in the preceding equation? If a negative sign or a subtraction symbol appears outside parentheses, drop the parentheses and change the sign of every term within the parentheses. For example,

(3x27x4)=3x2+7x+4.

Example 9 Simplifying an Algebraic Expression

Simplify: 8x+2[5(x3)].

Solution

A solution to simplify 8 x + 2 left bracket 5 minus left parenthesis x minus 3 right parenthesis right bracket.

Check Point 9

  • Simplify: 6+4[7(x2)].

Objective 10: Simplify algebraic expressions

Simplifying Algebraic Expressions

  1. Objective 10Simplify algebraic expressions.

Watch Video

The terms of an algebraic expression are those parts that are separated by addition. For example, consider the algebraic expression

7x9y+z3,

which can be expressed as

7x+(9y)+z+(3).

This expression contains four terms, namely, 7x, 9y, z, and 3.

The numerical part of a term is called its coefficient. In the term 7x, the 7 is the coefficient. If a term containing one or more variables is written without a coefficient, the coefficient is understood to be 1. Thus, z means 1z. If a term is a constant, its coefficient is that constant. Thus, the coefficient of the constant term 3 is 3.

7 x + left parenthesis negative 9 y right parenthesis + z + left parenthesis negative 3 right parenthesis.
P.1-22 Full Alternative Text

The parts of each term that are multiplied are called the factors of the term. The factors of the term 7x are 7 and x.

Like terms are terms that have exactly the same variable factors. For example, 3x and 7x are like terms. The distributive property in the form

ba+ca=(b+c)a

enables us to add or subtract like terms. For example,

3x+7x=(3+7)x=10x7y2y2=7y21y2=(71)y2= 6y2.

This process is called combining like terms.

An algebraic expression is simplified when parentheses have been removed and like terms have been combined.

Example 8 Simplifying an Algebraic Expression

Simplify: 6(2x2+4x)+10(4x2+3x).

Solution

Solution to simplify 6 left parenthesis 2 x squared + 4 x right parenthesis + 10 left parenthesis 4 x squared + 3 x right parenthesis. Arrows extend from 6 and 10 to the respective x squared and x terms.

Check Point 8

  • Simplify: 7(4x2+3x)+2(5x2+x).

Properties of Negatives

The distributive property can be extended to cover more than two terms within parentheses. For example,

A solution of negative 3 left parenthesis 4 x minus 2 y + 6 right parenthesis.
P.1-24 Full Alternative Text

The voice balloons illustrate that negative signs can appear side by side. They can represent the operation of subtraction or the fact that a real number is negative. Here is a list of properties of negatives and how they are applied to algebraic expressions:

Properties of Negatives

Let a and b represent real numbers, variables, or algebraic expressions.

Property Examples
1. (1)a=a (1)4xy=4xy
2. (a)=a (6y)=6y
3. (a)b=ab (7)4xy=74xy=28xy
4. a(b)=ab 5x(3y)=5x3y=15xy
5. (a+b)=ab (7x+6y)=7x6y

6. (ab)=a+b=ba

(3x7y)=3x+7y=7y3x

It is not uncommon to see algebraic expressions with parentheses preceded by a negative sign or subtraction. Properties 5 and 6 in the box, (a+b)=ab and (ab)=a+b, are related to this situation. An expression of the form (a+b) can be simplified as follows:

A polynomial equation.
P.1-25 Full Alternative Text

Do you see a fast way to obtain the simplified expression on the right in the preceding equation? If a negative sign or a subtraction symbol appears outside parentheses, drop the parentheses and change the sign of every term within the parentheses. For example,

(3x27x4)=3x2+7x+4.

Example 9 Simplifying an Algebraic Expression

Simplify: 8x+2[5(x3)].

Solution

A solution to simplify 8 x + 2 left bracket 5 minus left parenthesis x minus 3 right parenthesis right bracket.

Check Point 9

  • Simplify: 6+4[7(x2)].

P.1: Exercise Set

P.1 Exercise Set

Practice Exercises

In Exercises 116, evaluate each algebraic expression for the given value or values of the variable(s).

  1. 1. 7+5x, for x=10

  2. 2. 8+6x, for x=5

  3. 3. 6xy, for x=3 and y=8

  4. 4. 8xy, for x=3 and y=4

  5. 5. x2+3x, for x=8

  6. 6. x2+5x, for x=6

  7. 7. x26x+3, for x=7

  8. 8. x27x+4, for x=8

  9. 9. 4+5(x7)3, for x=9

  10. 10. 6+5(x6)3, for x=8

  11. 11. x23(xy), for x=8 and y=2

  12. 12. x24(xy), for x=8 and y=3

  13. 13. 5(x+2)2x14, for x=10

  14. 14. 7(x3)2x16, for x=9

  15. 15. 2x+3yx+1, for x=2 and y=4

  16. 16. 2x+yxy2x , for x=2 and y=4

The formula

C=59(F32)

expresses the relationship between Fahrenheit temperature, F, and Celsius temperature, C. In Exercises 1718, use the formula to convert the given Fahrenheit temperature to its equivalent temperature on the Celsius scale.

  1. 17. 50°F

  2. 18. 86°F

A football was kicked vertically upward from a height of 4 feet with an initial speed of 60 feet per second. The formula

h=4+60t16t2

describes the ball’s height above the ground, h, in feet, t seconds after it was kicked. Use this formula to solve Exercises 1920.

  1. 19. What was the ball’s height 2 seconds after it was kicked?

  2. 20. What was the ball’s height 3 seconds after it was kicked?

In Exercises 2128, find the intersection of the sets.

  1. 21. {1, 2, 3, 4}{2, 4, 5}

  2. 22. {1, 3, 7}{2, 3, 8}

  3. 23. {s, e, t}{t, e, s}

  4. 24. {r, e, a, l}{l, e, a, r}

  5. 25. {1, 3, 5, 7}{2, 4, 6, 8, 10}

  6. 26. {0, 1, 3, 5}{5, 3, 1}

  7. 27. {a, b, c, d}Ø

  8. 28. {w, y, z}Ø

In Exercises 2934, find the union of the sets.

  1. 29. {1, 2, 3, 4}{2, 4, 5}

  2. 30. {1, 3, 7, 8}{2, 3, 8}

  3. 31. {1, 3, 5, 7}{2, 4, 6, 8, 10}

  4. 32. {0, 1, 3, 5}{2, 4, 6}

  5. 33. {a, e, i, o, u}Ø

  6. 34. {e, m, p, t, y}Ø

In Exercises 3538, list all numbers from the given set that are a. natural numbers, b. whole numbers, c. integers, d. rational numbers, e. irrational numbers, f. real numbers.

  1. 35. {9, 45, 0, 0.25, 3 , 9.2, 100}

  2. 36. {7, 0.6¯, 0, 49, 50}

  3. 37. {11, 56, 0, 0.75, 5, π, 64}

  4. 38. {5, 0.3¯, 0, 2, 4}

  5. 39. Give an example of a whole number that is not a natural number.

  6. 40. Give an example of a rational number that is not an integer.

  7. 41. Give an example of a number that is an integer, a whole number, and a natural number.

  8. 42. Give an example of a number that is a rational number, an integer, and a real number.

Determine whether each statement in Exercises 4350 is true or false.

  1. 43. 132

  2. 44. 6>2

  3. 45. 47

  4. 46. 13<5

  5. 47. ππ

  6. 48. 3>13

  7. 49. 06

  8. 50. 013

In Exercises 5160, rewrite each expression without absolute value bars.

  1. 51. |300|

  2. 52. |203|

  3. 53. |12π|

  4. 54. |7π|

  5. 55. |25|

  6. 56. |513|

  7. 57. 3|3|

  8. 58. 7|7| 

  9. 59. 3||7

  10. 60. 5||13

In Exercises 6166, evaluate each algebraic expression for x=2 and y=5.

  1. 61. |x+y|

  2. 62. |xy|

  3. 63. |x|+|y|

  4. 64. |x||y|

  5. 65. y|y| 

  6. 66. |x|x+|y|y

In Exercises 6774, express the distance between the given numbers using absolute value. Then find the distance by evaluating the absolute value expression.

  1. 67. 2 and 17

  2. 68. 4 and 15

  3. 69. 2 and 5

  4. 70. 6 and 8

  5. 71. 19 and 4

  6. 72. 26 and 3

  7. 73. 3.6 and 1.4

  8. 74. 5.4 and 1.2

In Exercises 7584, state the name of the property illustrated.

  1. 75. 6+(4)=(4)+6

  2. 76. 11(7+4)=117+114

  3. 77. 6+(2+7)=(6+2)+7

  4. 78. 6(23)=6(32)

  5. 79. (2+3)+(4+5)=(4+5)+(2+3)

  6. 80. 7(118)=(118)7

  7. 81. 2(8+6)=16+12

  8. 82. 8(3+11)=24+(88)

  9. 83. 1(x+3)  (x+3)=1, x3

  10. 84. (x+4)+[(x+4)]=0

In Exercises 8596, simplify each algebraic expression.

  1. 85. 5(3x+4)4

  2. 86. 2(5x+4)3

  3. 87. 5(3x2)+12x

  4. 88. 2(5x1)+14x

  5. 89. 7(3y5)+2(4y+3)

  6. 90. 4(2y6)+3(5y+10)

  7. 91. 5(3y2)(7y+2)

  8. 92. 4(5y3)(6y+3)

  9. 93. 74[3(4y5)]

  10. 94. 65[8(2y4)]

  11. 95. 18x2+4[6(x22)+5]

  12. 96. 14x2+5[7(x22)+4]

In Exercises 97102, write each algebraic expression without parentheses.

  1. 97. (14x)

  2. 98. (17y)

  3. 99. (2x3y6)

  4. 100. (5x13y1)

  5. 101. 13(3x)+[(4y)+(4y)]

  6. 102. 12(2y)+[(7x)+7x]

Practice PLUS

In Exercises 103110, insert either <, >, or = in the shaded area to make a true statement.

  1. 103. |6| 10 |3|

  2. 104. |20| 10 |50|

  3. 105. |35| 00 |0.6|

  4. 106. |52 |00 |2.5|

  5. 107. 304034 00 14151514

  6. 108. 17181817 00 506056

  7. 109. 813÷813 00 |1|

  8. 110. |2| 00 417÷417

In Exercises 111120, use the order of operations to simplify each expression.

  1. 111. 8216÷2243

  2. 112. 102100÷5223

  3. 113. 5232[32(2)]2

  4. 114. 10÷2+34(1232)2

  5. 115. 83[2(25)4(86)]

  6. 116. 83[2(57)5(42)]

  7. 117. 2(2)4(3)58

  8. 118. 6(4)5(3)910

  9. 119. (56)22|37|89352

  10. 120. 12÷35|22+32|7+362

In Exercises 121128, write each English phrase as an algebraic expression. Then simplify the expression. Let x represent the number.

  1. 121. A number decreased by the sum of the number and four

  2. 122. A number decreased by the difference between eight and the number

  3. 123. Six times the product of negative five and a number

  4. 124. Ten times the product of negative four and a number

  5. 125. The difference between the product of five and a number and twice the number

  6. 126. The difference between the product of six and a number and negative two times the number

  7. 127. The difference between eight times a number and six more than three times the number

  8. 128. Eight decreased by three times the sum of a number and six

Application Exercises

The maximum heart rate, in beats per minute, that you should achieve during exercise is 220 minus your age:

220 minus a, is labeled, this algebraic expression gives maximum heart rate in terms of age, a.

The following graph shows the target heart rate ranges for four types of exercise goals. The lower and upper limits of these ranges are fractions of the maximum heart rate, 220a. Exercises 129130 are based on the information in the graph.

A horizontal bar graph titled, Target Heart Rate Ranges for Exercise Goals.
  1. 129. If your exercise goal is to improve cardiovascular conditioning, the graph shows the following range for target heart rate, H, in beats per minute:

    The lower limit of range labeled to, H = 7 tenths left parenthesis 200 minus a right parenthesis. The upper limit of range labeled to, H = 4 fifths left parenthesis 200 minus a right parenthesis.
    1. What is the lower limit of the heart rate range, in beats per minute, for a 20-year-old with this exercise goal?

    2. What is the upper limit of the heart rate range, in beats per minute, for a 20-year-old with this exercise goal?

  2. 130. If your exercise goal is to improve overall health, the graph shows the following range for target heart rate, H, in beats per minute:

    The lower limit of range labeled to, H = 1 half left parenthesis 200 minus a right parenthesis. The upper limit of range labeled to, H = 3 fifths left parenthesis 200 minus a right parenthesis.
    1. What is the lower limit of the heart rate range, in beats per minute, for a 30-year-old with this exercise goal?

    2. What is the upper limit of the heart rate range, in beats per minute, for a 30-year-old with this exercise goal?

The bar graph shows the average cost of tuition and fees at private four-year colleges in the United States.

A vertical bar graph titled, the average cost of tuition and fees at public four year U.S colleges.

Source: The College Board

Here are two formulas that model the data shown in the graph. In each formula, T represents the average cost of tuition and fees at private U.S. colleges for the school year ending x years after 2000.

Model 1 labeled to, T = 975 x + 13,547. Model 2 labeled to, T = 32 x squared + 331 x + 15,479.

Use this information to solve Exercises 131132.

  1. 131.

    1. Use each formula to find the average cost of tuition and fees at private U.S. colleges for the school year ending in 2018. By how much does each model underestimate or overestimate the actual cost shown for the school year ending in 2018?

    2. Use model 2 to project the average cost of tuition and fees at private U.S. colleges for the school year ending in 2030.

  2. 132.

    1. Use each formula to find the average cost of tuition and fees at private U.S. colleges for the school year ending in 2010. By how much does each underestimate or overestimate the actual cost shown for the school year ending in 2010?

    2. Use model 2 to project the average cost of tuition and fees at private U.S. colleges for the school year ending in 2025.

  3. 133. This month you have a total of $6000 in interest-bearing credit card debt, split between a card charging 18% annual interest and a card charging 21% annual interest. If the interest-bearing balance on the card charging 18% is x dollars, then the total interest for the month is given by the algebraic expression

    0.015x+0.0175(6000x).
    1. Simplify the algebraic expression.

    2. Use each form of the algebraic expression to determine the total interest for the month if the balance on the card charging 18% is $4400.

    3. Use the simplified form of the algebraic expression to determine the total interest for the month if the $6000 debt is split evenly between the two cards.

  4. 134. It takes you 50 minutes to get to campus. You spend t minutes walking to the bus stop and the rest of the time riding the bus. Your walking rate is 0.06 mile per minute and the bus travels at a rate of 0.5 mile per minute. The total distance walking and traveling by bus is given by the algebraic expression

    0.06t+0.5(50t).
    1. Simplify the algebraic expression.

    2. Use each form of the algebraic expression to determine the total distance that you travel if you spend 20 minutes walking to the bus stop.

    3. Use the simplified form of the algebraic expression to determine the total distance you travel if the 50 minutes is split evenly between walking and riding the bus.

  5. 135. Read the Blitzer Bonus beginning here. Use the formula

    BAC=600nw(0.6n+169)

    and replace w with your body weight. Using this formula and a calculator, compute your BAC for integers from n=1 to n=10. Round to three decimal places. According to this model, how many drinks can you consume in an hour without exceeding the legal measure of drunk driving?

Explaining the Concepts

  1. 136. What is an algebraic expression? Give an example with your explanation.

  2. 137. If n is a natural number, what does bn mean? Give an example with your explanation.

  3. 138. What does it mean when we say that a formula models real-world phenomena?

  4. 139. What is the intersection of sets A and B?

  5. 140. What is the union of sets A and B?

  6. 141. How do the whole numbers differ from the natural numbers?

  7. 142. Can a real number be both rational and irrational? Explain your answer.

  8. 143. If you are given two real numbers, explain how to determine which is the lesser.

Critical Thinking Exercises

Make Sense? In Exercises 144147, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 144. My mathematical model describes the data for tuition and fees at public four-year colleges for the past 20 years extremely well, so it will serve as an accurate prediction for the cost of public colleges in 2050.

  2. 145. A model that describes the average cost of tuition and fees at private U.S. colleges for the school year ending x years after 2000 cannot be used to estimate the cost of private education for the school year ending in 2000.

  3. 146. Regardless of what real numbers I substitute for x and y, I will always obtain zero when evaluating 2x2y2yx2.

  4. 147. Just as the commutative properties change groupings, the associative properties change order.

In Exercises 148155, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 148. Every rational number is an integer.

  2. 149. Some whole numbers are not integers.

  3. 150. Some rational numbers are not positive.

  4. 151. Irrational numbers cannot be negative.

  5. 152. The term x has no coefficient.

  6. 153. 5+3(x4)=8(x4)=8x32

  7. 154. xx=x+(x)=0

  8. 155. x0.02(x+200)=0.98x4

In Exercises 156158, insert either < or > in the shaded area between the numbers to make the statement true.

  1. 156. 2 00 1.5

  2. 157. π 00 3.5

  3. 158. 3.142 00 π2

Preview Exercises

Exercises 159161 will help you prepare for the material covered in the next section.

  1. 159. In parts (a) and (b), complete each statement.

    1. b4b3=(bbbb)(bbb)=b?

    2. b5b5=(bbbbb)(bbbbb)=b?

    3. Generalizing from parts (a) and (b), what should be done with the exponents when multiplying exponential expressions with the same base?

  2. 160. In parts (a) and (b), complete each statement.

    1. b7b3=bbbbbbbbb b=b?

    2. b8b2=bbbbbbbbbb =b?

    3. Generalizing from parts (a) and (b), what should be done with the exponents when dividing exponential expressions with the same base?

  3. 161. If 6.2 is multiplied by 103, what does this multiplication do to the decimal point in 6.2?

P.1: Exercise Set

P.1 Exercise Set

Practice Exercises

In Exercises 116, evaluate each algebraic expression for the given value or values of the variable(s).

  1. 1. 7+5x, for x=10

  2. 2. 8+6x, for x=5

  3. 3. 6xy, for x=3 and y=8

  4. 4. 8xy, for x=3 and y=4

  5. 5. x2+3x, for x=8

  6. 6. x2+5x, for x=6

  7. 7. x26x+3, for x=7

  8. 8. x27x+4, for x=8

  9. 9. 4+5(x7)3, for x=9

  10. 10. 6+5(x6)3, for x=8

  11. 11. x23(xy), for x=8 and y=2

  12. 12. x24(xy), for x=8 and y=3

  13. 13. 5(x+2)2x14, for x=10

  14. 14. 7(x3)2x16, for x=9

  15. 15. 2x+3yx+1, for x=2 and y=4

  16. 16. 2x+yxy2x , for x=2 and y=4

The formula

C=59(F32)

expresses the relationship between Fahrenheit temperature, F, and Celsius temperature, C. In Exercises 1718, use the formula to convert the given Fahrenheit temperature to its equivalent temperature on the Celsius scale.

  1. 17. 50°F

  2. 18. 86°F

A football was kicked vertically upward from a height of 4 feet with an initial speed of 60 feet per second. The formula

h=4+60t16t2

describes the ball’s height above the ground, h, in feet, t seconds after it was kicked. Use this formula to solve Exercises 1920.

  1. 19. What was the ball’s height 2 seconds after it was kicked?

  2. 20. What was the ball’s height 3 seconds after it was kicked?

In Exercises 2128, find the intersection of the sets.

  1. 21. {1, 2, 3, 4}{2, 4, 5}

  2. 22. {1, 3, 7}{2, 3, 8}

  3. 23. {s, e, t}{t, e, s}

  4. 24. {r, e, a, l}{l, e, a, r}

  5. 25. {1, 3, 5, 7}{2, 4, 6, 8, 10}

  6. 26. {0, 1, 3, 5}{5, 3, 1}

  7. 27. {a, b, c, d}Ø

  8. 28. {w, y, z}Ø

In Exercises 2934, find the union of the sets.

  1. 29. {1, 2, 3, 4}{2, 4, 5}

  2. 30. {1, 3, 7, 8}{2, 3, 8}

  3. 31. {1, 3, 5, 7}{2, 4, 6, 8, 10}

  4. 32. {0, 1, 3, 5}{2, 4, 6}

  5. 33. {a, e, i, o, u}Ø

  6. 34. {e, m, p, t, y}Ø

In Exercises 3538, list all numbers from the given set that are a. natural numbers, b. whole numbers, c. integers, d. rational numbers, e. irrational numbers, f. real numbers.

  1. 35. {9, 45, 0, 0.25, 3 , 9.2, 100}

  2. 36. {7, 0.6¯, 0, 49, 50}

  3. 37. {11, 56, 0, 0.75, 5, π, 64}

  4. 38. {5, 0.3¯, 0, 2, 4}

  5. 39. Give an example of a whole number that is not a natural number.

  6. 40. Give an example of a rational number that is not an integer.

  7. 41. Give an example of a number that is an integer, a whole number, and a natural number.

  8. 42. Give an example of a number that is a rational number, an integer, and a real number.

Determine whether each statement in Exercises 4350 is true or false.

  1. 43. 132

  2. 44. 6>2

  3. 45. 47

  4. 46. 13<5

  5. 47. ππ

  6. 48. 3>13

  7. 49. 06

  8. 50. 013

In Exercises 5160, rewrite each expression without absolute value bars.

  1. 51. |300|

  2. 52. |203|

  3. 53. |12π|

  4. 54. |7π|

  5. 55. |25|

  6. 56. |513|

  7. 57. 3|3|

  8. 58. 7|7| 

  9. 59. 3||7

  10. 60. 5||13

In Exercises 6166, evaluate each algebraic expression for x=2 and y=5.

  1. 61. |x+y|

  2. 62. |xy|

  3. 63. |x|+|y|

  4. 64. |x||y|

  5. 65. y|y| 

  6. 66. |x|x+|y|y

In Exercises 6774, express the distance between the given numbers using absolute value. Then find the distance by evaluating the absolute value expression.

  1. 67. 2 and 17

  2. 68. 4 and 15

  3. 69. 2 and 5

  4. 70. 6 and 8

  5. 71. 19 and 4

  6. 72. 26 and 3

  7. 73. 3.6 and 1.4

  8. 74. 5.4 and 1.2

In Exercises 7584, state the name of the property illustrated.

  1. 75. 6+(4)=(4)+6

  2. 76. 11(7+4)=117+114

  3. 77. 6+(2+7)=(6+2)+7

  4. 78. 6(23)=6(32)

  5. 79. (2+3)+(4+5)=(4+5)+(2+3)

  6. 80. 7(118)=(118)7

  7. 81. 2(8+6)=16+12

  8. 82. 8(3+11)=24+(88)

  9. 83. 1(x+3)  (x+3)=1, x3

  10. 84. (x+4)+[(x+4)]=0

In Exercises 8596, simplify each algebraic expression.

  1. 85. 5(3x+4)4

  2. 86. 2(5x+4)3

  3. 87. 5(3x2)+12x

  4. 88. 2(5x1)+14x

  5. 89. 7(3y5)+2(4y+3)

  6. 90. 4(2y6)+3(5y+10)

  7. 91. 5(3y2)(7y+2)

  8. 92. 4(5y3)(6y+3)

  9. 93. 74[3(4y5)]

  10. 94. 65[8(2y4)]

  11. 95. 18x2+4[6(x22)+5]

  12. 96. 14x2+5[7(x22)+4]

In Exercises 97102, write each algebraic expression without parentheses.

  1. 97. (14x)

  2. 98. (17y)

  3. 99. (2x3y6)

  4. 100. (5x13y1)

  5. 101. 13(3x)+[(4y)+(4y)]

  6. 102. 12(2y)+[(7x)+7x]

Practice PLUS

In Exercises 103110, insert either <, >, or = in the shaded area to make a true statement.

  1. 103. |6| 10 |3|

  2. 104. |20| 10 |50|

  3. 105. |35| 00 |0.6|

  4. 106. |52 |00 |2.5|

  5. 107. 304034 00 14151514

  6. 108. 17181817 00 506056

  7. 109. 813÷813 00 |1|

  8. 110. |2| 00 417÷417

In Exercises 111120, use the order of operations to simplify each expression.

  1. 111. 8216÷2243

  2. 112. 102100÷5223

  3. 113. 5232[32(2)]2

  4. 114. 10÷2+34(1232)2

  5. 115. 83[2(25)4(86)]

  6. 116. 83[2(57)5(42)]

  7. 117. 2(2)4(3)58

  8. 118. 6(4)5(3)910

  9. 119. (56)22|37|89352

  10. 120. 12÷35|22+32|7+362

In Exercises 121128, write each English phrase as an algebraic expression. Then simplify the expression. Let x represent the number.

  1. 121. A number decreased by the sum of the number and four

  2. 122. A number decreased by the difference between eight and the number

  3. 123. Six times the product of negative five and a number

  4. 124. Ten times the product of negative four and a number

  5. 125. The difference between the product of five and a number and twice the number

  6. 126. The difference between the product of six and a number and negative two times the number

  7. 127. The difference between eight times a number and six more than three times the number

  8. 128. Eight decreased by three times the sum of a number and six

Application Exercises

The maximum heart rate, in beats per minute, that you should achieve during exercise is 220 minus your age:

220 minus a, is labeled, this algebraic expression gives maximum heart rate in terms of age, a.

The following graph shows the target heart rate ranges for four types of exercise goals. The lower and upper limits of these ranges are fractions of the maximum heart rate, 220a. Exercises 129130 are based on the information in the graph.

A horizontal bar graph titled, Target Heart Rate Ranges for Exercise Goals.
  1. 129. If your exercise goal is to improve cardiovascular conditioning, the graph shows the following range for target heart rate, H, in beats per minute:

    The lower limit of range labeled to, H = 7 tenths left parenthesis 200 minus a right parenthesis. The upper limit of range labeled to, H = 4 fifths left parenthesis 200 minus a right parenthesis.
    1. What is the lower limit of the heart rate range, in beats per minute, for a 20-year-old with this exercise goal?

    2. What is the upper limit of the heart rate range, in beats per minute, for a 20-year-old with this exercise goal?

  2. 130. If your exercise goal is to improve overall health, the graph shows the following range for target heart rate, H, in beats per minute:

    The lower limit of range labeled to, H = 1 half left parenthesis 200 minus a right parenthesis. The upper limit of range labeled to, H = 3 fifths left parenthesis 200 minus a right parenthesis.
    1. What is the lower limit of the heart rate range, in beats per minute, for a 30-year-old with this exercise goal?

    2. What is the upper limit of the heart rate range, in beats per minute, for a 30-year-old with this exercise goal?

The bar graph shows the average cost of tuition and fees at private four-year colleges in the United States.

A vertical bar graph titled, the average cost of tuition and fees at public four year U.S colleges.

Source: The College Board

Here are two formulas that model the data shown in the graph. In each formula, T represents the average cost of tuition and fees at private U.S. colleges for the school year ending x years after 2000.

Model 1 labeled to, T = 975 x + 13,547. Model 2 labeled to, T = 32 x squared + 331 x + 15,479.

Use this information to solve Exercises 131132.

  1. 131.

    1. Use each formula to find the average cost of tuition and fees at private U.S. colleges for the school year ending in 2018. By how much does each model underestimate or overestimate the actual cost shown for the school year ending in 2018?

    2. Use model 2 to project the average cost of tuition and fees at private U.S. colleges for the school year ending in 2030.

  2. 132.

    1. Use each formula to find the average cost of tuition and fees at private U.S. colleges for the school year ending in 2010. By how much does each underestimate or overestimate the actual cost shown for the school year ending in 2010?

    2. Use model 2 to project the average cost of tuition and fees at private U.S. colleges for the school year ending in 2025.

  3. 133. This month you have a total of $6000 in interest-bearing credit card debt, split between a card charging 18% annual interest and a card charging 21% annual interest. If the interest-bearing balance on the card charging 18% is x dollars, then the total interest for the month is given by the algebraic expression

    0.015x+0.0175(6000x).
    1. Simplify the algebraic expression.

    2. Use each form of the algebraic expression to determine the total interest for the month if the balance on the card charging 18% is $4400.

    3. Use the simplified form of the algebraic expression to determine the total interest for the month if the $6000 debt is split evenly between the two cards.

  4. 134. It takes you 50 minutes to get to campus. You spend t minutes walking to the bus stop and the rest of the time riding the bus. Your walking rate is 0.06 mile per minute and the bus travels at a rate of 0.5 mile per minute. The total distance walking and traveling by bus is given by the algebraic expression

    0.06t+0.5(50t).
    1. Simplify the algebraic expression.

    2. Use each form of the algebraic expression to determine the total distance that you travel if you spend 20 minutes walking to the bus stop.

    3. Use the simplified form of the algebraic expression to determine the total distance you travel if the 50 minutes is split evenly between walking and riding the bus.

  5. 135. Read the Blitzer Bonus beginning here. Use the formula

    BAC=600nw(0.6n+169)

    and replace w with your body weight. Using this formula and a calculator, compute your BAC for integers from n=1 to n=10. Round to three decimal places. According to this model, how many drinks can you consume in an hour without exceeding the legal measure of drunk driving?

Explaining the Concepts

  1. 136. What is an algebraic expression? Give an example with your explanation.

  2. 137. If n is a natural number, what does bn mean? Give an example with your explanation.

  3. 138. What does it mean when we say that a formula models real-world phenomena?

  4. 139. What is the intersection of sets A and B?

  5. 140. What is the union of sets A and B?

  6. 141. How do the whole numbers differ from the natural numbers?

  7. 142. Can a real number be both rational and irrational? Explain your answer.

  8. 143. If you are given two real numbers, explain how to determine which is the lesser.

Critical Thinking Exercises

Make Sense? In Exercises 144147, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 144. My mathematical model describes the data for tuition and fees at public four-year colleges for the past 20 years extremely well, so it will serve as an accurate prediction for the cost of public colleges in 2050.

  2. 145. A model that describes the average cost of tuition and fees at private U.S. colleges for the school year ending x years after 2000 cannot be used to estimate the cost of private education for the school year ending in 2000.

  3. 146. Regardless of what real numbers I substitute for x and y, I will always obtain zero when evaluating 2x2y2yx2.

  4. 147. Just as the commutative properties change groupings, the associative properties change order.

In Exercises 148155, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 148. Every rational number is an integer.

  2. 149. Some whole numbers are not integers.

  3. 150. Some rational numbers are not positive.

  4. 151. Irrational numbers cannot be negative.

  5. 152. The term x has no coefficient.

  6. 153. 5+3(x4)=8(x4)=8x32

  7. 154. xx=x+(x)=0

  8. 155. x0.02(x+200)=0.98x4

In Exercises 156158, insert either < or > in the shaded area between the numbers to make the statement true.

  1. 156. 2 00 1.5

  2. 157. π 00 3.5

  3. 158. 3.142 00 π2

Preview Exercises

Exercises 159161 will help you prepare for the material covered in the next section.

  1. 159. In parts (a) and (b), complete each statement.

    1. b4b3=(bbbb)(bbb)=b?

    2. b5b5=(bbbbb)(bbbbb)=b?

    3. Generalizing from parts (a) and (b), what should be done with the exponents when multiplying exponential expressions with the same base?

  2. 160. In parts (a) and (b), complete each statement.

    1. b7b3=bbbbbbbbb b=b?

    2. b8b2=bbbbbbbbbb =b?

    3. Generalizing from parts (a) and (b), what should be done with the exponents when dividing exponential expressions with the same base?

  3. 161. If 6.2 is multiplied by 103, what does this multiplication do to the decimal point in 6.2?

P.1: Exercise Set

P.1 Exercise Set

Practice Exercises

In Exercises 116, evaluate each algebraic expression for the given value or values of the variable(s).

  1. 1. 7+5x, for x=10

  2. 2. 8+6x, for x=5

  3. 3. 6xy, for x=3 and y=8

  4. 4. 8xy, for x=3 and y=4

  5. 5. x2+3x, for x=8

  6. 6. x2+5x, for x=6

  7. 7. x26x+3, for x=7

  8. 8. x27x+4, for x=8

  9. 9. 4+5(x7)3, for x=9

  10. 10. 6+5(x6)3, for x=8

  11. 11. x23(xy), for x=8 and y=2

  12. 12. x24(xy), for x=8 and y=3

  13. 13. 5(x+2)2x14, for x=10

  14. 14. 7(x3)2x16, for x=9

  15. 15. 2x+3yx+1, for x=2 and y=4

  16. 16. 2x+yxy2x , for x=2 and y=4

The formula

C=59(F32)

expresses the relationship between Fahrenheit temperature, F, and Celsius temperature, C. In Exercises 1718, use the formula to convert the given Fahrenheit temperature to its equivalent temperature on the Celsius scale.

  1. 17. 50°F

  2. 18. 86°F

A football was kicked vertically upward from a height of 4 feet with an initial speed of 60 feet per second. The formula

h=4+60t16t2

describes the ball’s height above the ground, h, in feet, t seconds after it was kicked. Use this formula to solve Exercises 1920.

  1. 19. What was the ball’s height 2 seconds after it was kicked?

  2. 20. What was the ball’s height 3 seconds after it was kicked?

In Exercises 2128, find the intersection of the sets.

  1. 21. {1, 2, 3, 4}{2, 4, 5}

  2. 22. {1, 3, 7}{2, 3, 8}

  3. 23. {s, e, t}{t, e, s}

  4. 24. {r, e, a, l}{l, e, a, r}

  5. 25. {1, 3, 5, 7}{2, 4, 6, 8, 10}

  6. 26. {0, 1, 3, 5}{5, 3, 1}

  7. 27. {a, b, c, d}Ø

  8. 28. {w, y, z}Ø

In Exercises 2934, find the union of the sets.

  1. 29. {1, 2, 3, 4}{2, 4, 5}

  2. 30. {1, 3, 7, 8}{2, 3, 8}

  3. 31. {1, 3, 5, 7}{2, 4, 6, 8, 10}

  4. 32. {0, 1, 3, 5}{2, 4, 6}

  5. 33. {a, e, i, o, u}Ø

  6. 34. {e, m, p, t, y}Ø

In Exercises 3538, list all numbers from the given set that are a. natural numbers, b. whole numbers, c. integers, d. rational numbers, e. irrational numbers, f. real numbers.

  1. 35. {9, 45, 0, 0.25, 3 , 9.2, 100}

  2. 36. {7, 0.6¯, 0, 49, 50}

  3. 37. {11, 56, 0, 0.75, 5, π, 64}

  4. 38. {5, 0.3¯, 0, 2, 4}

  5. 39. Give an example of a whole number that is not a natural number.

  6. 40. Give an example of a rational number that is not an integer.

  7. 41. Give an example of a number that is an integer, a whole number, and a natural number.

  8. 42. Give an example of a number that is a rational number, an integer, and a real number.

Determine whether each statement in Exercises 4350 is true or false.

  1. 43. 132

  2. 44. 6>2

  3. 45. 47

  4. 46. 13<5

  5. 47. ππ

  6. 48. 3>13

  7. 49. 06

  8. 50. 013

In Exercises 5160, rewrite each expression without absolute value bars.

  1. 51. |300|

  2. 52. |203|

  3. 53. |12π|

  4. 54. |7π|

  5. 55. |25|

  6. 56. |513|

  7. 57. 3|3|

  8. 58. 7|7| 

  9. 59. 3||7

  10. 60. 5||13

In Exercises 6166, evaluate each algebraic expression for x=2 and y=5.

  1. 61. |x+y|

  2. 62. |xy|

  3. 63. |x|+|y|

  4. 64. |x||y|

  5. 65. y|y| 

  6. 66. |x|x+|y|y

In Exercises 6774, express the distance between the given numbers using absolute value. Then find the distance by evaluating the absolute value expression.

  1. 67. 2 and 17

  2. 68. 4 and 15

  3. 69. 2 and 5

  4. 70. 6 and 8

  5. 71. 19 and 4

  6. 72. 26 and 3

  7. 73. 3.6 and 1.4

  8. 74. 5.4 and 1.2

In Exercises 7584, state the name of the property illustrated.

  1. 75. 6+(4)=(4)+6

  2. 76. 11(7+4)=117+114

  3. 77. 6+(2+7)=(6+2)+7

  4. 78. 6(23)=6(32)

  5. 79. (2+3)+(4+5)=(4+5)+(2+3)

  6. 80. 7(118)=(118)7

  7. 81. 2(8+6)=16+12

  8. 82. 8(3+11)=24+(88)

  9. 83. 1(x+3)  (x+3)=1, x3

  10. 84. (x+4)+[(x+4)]=0

In Exercises 8596, simplify each algebraic expression.

  1. 85. 5(3x+4)4

  2. 86. 2(5x+4)3

  3. 87. 5(3x2)+12x

  4. 88. 2(5x1)+14x

  5. 89. 7(3y5)+2(4y+3)

  6. 90. 4(2y6)+3(5y+10)

  7. 91. 5(3y2)(7y+2)

  8. 92. 4(5y3)(6y+3)

  9. 93. 74[3(4y5)]

  10. 94. 65[8(2y4)]

  11. 95. 18x2+4[6(x22)+5]

  12. 96. 14x2+5[7(x22)+4]

In Exercises 97102, write each algebraic expression without parentheses.

  1. 97. (14x)

  2. 98. (17y)

  3. 99. (2x3y6)

  4. 100. (5x13y1)

  5. 101. 13(3x)+[(4y)+(4y)]

  6. 102. 12(2y)+[(7x)+7x]

Practice PLUS

In Exercises 103110, insert either <, >, or = in the shaded area to make a true statement.

  1. 103. |6| 10 |3|

  2. 104. |20| 10 |50|

  3. 105. |35| 00 |0.6|

  4. 106. |52 |00 |2.5|

  5. 107. 304034 00 14151514

  6. 108. 17181817 00 506056

  7. 109. 813÷813 00 |1|

  8. 110. |2| 00 417÷417

In Exercises 111120, use the order of operations to simplify each expression.

  1. 111. 8216÷2243

  2. 112. 102100÷5223

  3. 113. 5232[32(2)]2

  4. 114. 10÷2+34(1232)2

  5. 115. 83[2(25)4(86)]

  6. 116. 83[2(57)5(42)]

  7. 117. 2(2)4(3)58

  8. 118. 6(4)5(3)910

  9. 119. (56)22|37|89352

  10. 120. 12÷35|22+32|7+362

In Exercises 121128, write each English phrase as an algebraic expression. Then simplify the expression. Let x represent the number.

  1. 121. A number decreased by the sum of the number and four

  2. 122. A number decreased by the difference between eight and the number

  3. 123. Six times the product of negative five and a number

  4. 124. Ten times the product of negative four and a number

  5. 125. The difference between the product of five and a number and twice the number

  6. 126. The difference between the product of six and a number and negative two times the number

  7. 127. The difference between eight times a number and six more than three times the number

  8. 128. Eight decreased by three times the sum of a number and six

Application Exercises

The maximum heart rate, in beats per minute, that you should achieve during exercise is 220 minus your age:

220 minus a, is labeled, this algebraic expression gives maximum heart rate in terms of age, a.

The following graph shows the target heart rate ranges for four types of exercise goals. The lower and upper limits of these ranges are fractions of the maximum heart rate, 220a. Exercises 129130 are based on the information in the graph.

A horizontal bar graph titled, Target Heart Rate Ranges for Exercise Goals.
  1. 129. If your exercise goal is to improve cardiovascular conditioning, the graph shows the following range for target heart rate, H, in beats per minute:

    The lower limit of range labeled to, H = 7 tenths left parenthesis 200 minus a right parenthesis. The upper limit of range labeled to, H = 4 fifths left parenthesis 200 minus a right parenthesis.
    1. What is the lower limit of the heart rate range, in beats per minute, for a 20-year-old with this exercise goal?

    2. What is the upper limit of the heart rate range, in beats per minute, for a 20-year-old with this exercise goal?

  2. 130. If your exercise goal is to improve overall health, the graph shows the following range for target heart rate, H, in beats per minute:

    The lower limit of range labeled to, H = 1 half left parenthesis 200 minus a right parenthesis. The upper limit of range labeled to, H = 3 fifths left parenthesis 200 minus a right parenthesis.
    1. What is the lower limit of the heart rate range, in beats per minute, for a 30-year-old with this exercise goal?

    2. What is the upper limit of the heart rate range, in beats per minute, for a 30-year-old with this exercise goal?

The bar graph shows the average cost of tuition and fees at private four-year colleges in the United States.

A vertical bar graph titled, the average cost of tuition and fees at public four year U.S colleges.

Source: The College Board

Here are two formulas that model the data shown in the graph. In each formula, T represents the average cost of tuition and fees at private U.S. colleges for the school year ending x years after 2000.

Model 1 labeled to, T = 975 x + 13,547. Model 2 labeled to, T = 32 x squared + 331 x + 15,479.

Use this information to solve Exercises 131132.

  1. 131.

    1. Use each formula to find the average cost of tuition and fees at private U.S. colleges for the school year ending in 2018. By how much does each model underestimate or overestimate the actual cost shown for the school year ending in 2018?

    2. Use model 2 to project the average cost of tuition and fees at private U.S. colleges for the school year ending in 2030.

  2. 132.

    1. Use each formula to find the average cost of tuition and fees at private U.S. colleges for the school year ending in 2010. By how much does each underestimate or overestimate the actual cost shown for the school year ending in 2010?

    2. Use model 2 to project the average cost of tuition and fees at private U.S. colleges for the school year ending in 2025.

  3. 133. This month you have a total of $6000 in interest-bearing credit card debt, split between a card charging 18% annual interest and a card charging 21% annual interest. If the interest-bearing balance on the card charging 18% is x dollars, then the total interest for the month is given by the algebraic expression

    0.015x+0.0175(6000x).
    1. Simplify the algebraic expression.

    2. Use each form of the algebraic expression to determine the total interest for the month if the balance on the card charging 18% is $4400.

    3. Use the simplified form of the algebraic expression to determine the total interest for the month if the $6000 debt is split evenly between the two cards.

  4. 134. It takes you 50 minutes to get to campus. You spend t minutes walking to the bus stop and the rest of the time riding the bus. Your walking rate is 0.06 mile per minute and the bus travels at a rate of 0.5 mile per minute. The total distance walking and traveling by bus is given by the algebraic expression

    0.06t+0.5(50t).
    1. Simplify the algebraic expression.

    2. Use each form of the algebraic expression to determine the total distance that you travel if you spend 20 minutes walking to the bus stop.

    3. Use the simplified form of the algebraic expression to determine the total distance you travel if the 50 minutes is split evenly between walking and riding the bus.

  5. 135. Read the Blitzer Bonus beginning here. Use the formula

    BAC=600nw(0.6n+169)

    and replace w with your body weight. Using this formula and a calculator, compute your BAC for integers from n=1 to n=10. Round to three decimal places. According to this model, how many drinks can you consume in an hour without exceeding the legal measure of drunk driving?

Explaining the Concepts

  1. 136. What is an algebraic expression? Give an example with your explanation.

  2. 137. If n is a natural number, what does bn mean? Give an example with your explanation.

  3. 138. What does it mean when we say that a formula models real-world phenomena?

  4. 139. What is the intersection of sets A and B?

  5. 140. What is the union of sets A and B?

  6. 141. How do the whole numbers differ from the natural numbers?

  7. 142. Can a real number be both rational and irrational? Explain your answer.

  8. 143. If you are given two real numbers, explain how to determine which is the lesser.

Critical Thinking Exercises

Make Sense? In Exercises 144147, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 144. My mathematical model describes the data for tuition and fees at public four-year colleges for the past 20 years extremely well, so it will serve as an accurate prediction for the cost of public colleges in 2050.

  2. 145. A model that describes the average cost of tuition and fees at private U.S. colleges for the school year ending x years after 2000 cannot be used to estimate the cost of private education for the school year ending in 2000.

  3. 146. Regardless of what real numbers I substitute for x and y, I will always obtain zero when evaluating 2x2y2yx2.

  4. 147. Just as the commutative properties change groupings, the associative properties change order.

In Exercises 148155, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 148. Every rational number is an integer.

  2. 149. Some whole numbers are not integers.

  3. 150. Some rational numbers are not positive.

  4. 151. Irrational numbers cannot be negative.

  5. 152. The term x has no coefficient.

  6. 153. 5+3(x4)=8(x4)=8x32

  7. 154. xx=x+(x)=0

  8. 155. x0.02(x+200)=0.98x4

In Exercises 156158, insert either < or > in the shaded area between the numbers to make the statement true.

  1. 156. 2 00 1.5

  2. 157. π 00 3.5

  3. 158. 3.142 00 π2

Preview Exercises

Exercises 159161 will help you prepare for the material covered in the next section.

  1. 159. In parts (a) and (b), complete each statement.

    1. b4b3=(bbbb)(bbb)=b?

    2. b5b5=(bbbbb)(bbbbb)=b?

    3. Generalizing from parts (a) and (b), what should be done with the exponents when multiplying exponential expressions with the same base?

  2. 160. In parts (a) and (b), complete each statement.

    1. b7b3=bbbbbbbbb b=b?

    2. b8b2=bbbbbbbbbb =b?

    3. Generalizing from parts (a) and (b), what should be done with the exponents when dividing exponential expressions with the same base?

  3. 161. If 6.2 is multiplied by 103, what does this multiplication do to the decimal point in 6.2?

P.1: Exercise Set

P.1 Exercise Set

Practice Exercises

In Exercises 116, evaluate each algebraic expression for the given value or values of the variable(s).

  1. 1. 7+5x, for x=10

  2. 2. 8+6x, for x=5

  3. 3. 6xy, for x=3 and y=8

  4. 4. 8xy, for x=3 and y=4

  5. 5. x2+3x, for x=8

  6. 6. x2+5x, for x=6

  7. 7. x26x+3, for x=7

  8. 8. x27x+4, for x=8

  9. 9. 4+5(x7)3, for x=9

  10. 10. 6+5(x6)3, for x=8

  11. 11. x23(xy), for x=8 and y=2

  12. 12. x24(xy), for x=8 and y=3

  13. 13. 5(x+2)2x14, for x=10

  14. 14. 7(x3)2x16, for x=9

  15. 15. 2x+3yx+1, for x=2 and y=4

  16. 16. 2x+yxy2x , for x=2 and y=4

The formula

C=59(F32)

expresses the relationship between Fahrenheit temperature, F, and Celsius temperature, C. In Exercises 1718, use the formula to convert the given Fahrenheit temperature to its equivalent temperature on the Celsius scale.

  1. 17. 50°F

  2. 18. 86°F

A football was kicked vertically upward from a height of 4 feet with an initial speed of 60 feet per second. The formula

h=4+60t16t2

describes the ball’s height above the ground, h, in feet, t seconds after it was kicked. Use this formula to solve Exercises 1920.

  1. 19. What was the ball’s height 2 seconds after it was kicked?

  2. 20. What was the ball’s height 3 seconds after it was kicked?

In Exercises 2128, find the intersection of the sets.

  1. 21. {1, 2, 3, 4}{2, 4, 5}

  2. 22. {1, 3, 7}{2, 3, 8}

  3. 23. {s, e, t}{t, e, s}

  4. 24. {r, e, a, l}{l, e, a, r}

  5. 25. {1, 3, 5, 7}{2, 4, 6, 8, 10}

  6. 26. {0, 1, 3, 5}{5, 3, 1}

  7. 27. {a, b, c, d}Ø

  8. 28. {w, y, z}Ø

In Exercises 2934, find the union of the sets.

  1. 29. {1, 2, 3, 4}{2, 4, 5}

  2. 30. {1, 3, 7, 8}{2, 3, 8}

  3. 31. {1, 3, 5, 7}{2, 4, 6, 8, 10}

  4. 32. {0, 1, 3, 5}{2, 4, 6}

  5. 33. {a, e, i, o, u}Ø

  6. 34. {e, m, p, t, y}Ø

In Exercises 3538, list all numbers from the given set that are a. natural numbers, b. whole numbers, c. integers, d. rational numbers, e. irrational numbers, f. real numbers.

  1. 35. {9, 45, 0, 0.25, 3 , 9.2, 100}

  2. 36. {7, 0.6¯, 0, 49, 50}

  3. 37. {11, 56, 0, 0.75, 5, π, 64}

  4. 38. {5, 0.3¯, 0, 2, 4}

  5. 39. Give an example of a whole number that is not a natural number.

  6. 40. Give an example of a rational number that is not an integer.

  7. 41. Give an example of a number that is an integer, a whole number, and a natural number.

  8. 42. Give an example of a number that is a rational number, an integer, and a real number.

Determine whether each statement in Exercises 4350 is true or false.

  1. 43. 132

  2. 44. 6>2

  3. 45. 47

  4. 46. 13<5

  5. 47. ππ

  6. 48. 3>13

  7. 49. 06

  8. 50. 013

In Exercises 5160, rewrite each expression without absolute value bars.

  1. 51. |300|

  2. 52. |203|

  3. 53. |12π|

  4. 54. |7π|

  5. 55. |25|

  6. 56. |513|

  7. 57. 3|3|

  8. 58. 7|7| 

  9. 59. 3||7

  10. 60. 5||13

In Exercises 6166, evaluate each algebraic expression for x=2 and y=5.

  1. 61. |x+y|

  2. 62. |xy|

  3. 63. |x|+|y|

  4. 64. |x||y|

  5. 65. y|y| 

  6. 66. |x|x+|y|y

In Exercises 6774, express the distance between the given numbers using absolute value. Then find the distance by evaluating the absolute value expression.

  1. 67. 2 and 17

  2. 68. 4 and 15

  3. 69. 2 and 5

  4. 70. 6 and 8

  5. 71. 19 and 4

  6. 72. 26 and 3

  7. 73. 3.6 and 1.4

  8. 74. 5.4 and 1.2

In Exercises 7584, state the name of the property illustrated.

  1. 75. 6+(4)=(4)+6

  2. 76. 11(7+4)=117+114

  3. 77. 6+(2+7)=(6+2)+7

  4. 78. 6(23)=6(32)

  5. 79. (2+3)+(4+5)=(4+5)+(2+3)

  6. 80. 7(118)=(118)7

  7. 81. 2(8+6)=16+12

  8. 82. 8(3+11)=24+(88)

  9. 83. 1(x+3)  (x+3)=1, x3

  10. 84. (x+4)+[(x+4)]=0

In Exercises 8596, simplify each algebraic expression.

  1. 85. 5(3x+4)4

  2. 86. 2(5x+4)3

  3. 87. 5(3x2)+12x

  4. 88. 2(5x1)+14x

  5. 89. 7(3y5)+2(4y+3)

  6. 90. 4(2y6)+3(5y+10)

  7. 91. 5(3y2)(7y+2)

  8. 92. 4(5y3)(6y+3)

  9. 93. 74[3(4y5)]

  10. 94. 65[8(2y4)]

  11. 95. 18x2+4[6(x22)+5]

  12. 96. 14x2+5[7(x22)+4]

In Exercises 97102, write each algebraic expression without parentheses.

  1. 97. (14x)

  2. 98. (17y)

  3. 99. (2x3y6)

  4. 100. (5x13y1)

  5. 101. 13(3x)+[(4y)+(4y)]

  6. 102. 12(2y)+[(7x)+7x]

Practice PLUS

In Exercises 103110, insert either <, >, or = in the shaded area to make a true statement.

  1. 103. |6| 10 |3|

  2. 104. |20| 10 |50|

  3. 105. |35| 00 |0.6|

  4. 106. |52 |00 |2.5|

  5. 107. 304034 00 14151514

  6. 108. 17181817 00 506056

  7. 109. 813÷813 00 |1|

  8. 110. |2| 00 417÷417

In Exercises 111120, use the order of operations to simplify each expression.

  1. 111. 8216÷2243

  2. 112. 102100÷5223

  3. 113. 5232[32(2)]2

  4. 114. 10÷2+34(1232)2

  5. 115. 83[2(25)4(86)]

  6. 116. 83[2(57)5(42)]

  7. 117. 2(2)4(3)58

  8. 118. 6(4)5(3)910

  9. 119. (56)22|37|89352

  10. 120. 12÷35|22+32|7+362

In Exercises 121128, write each English phrase as an algebraic expression. Then simplify the expression. Let x represent the number.

  1. 121. A number decreased by the sum of the number and four

  2. 122. A number decreased by the difference between eight and the number

  3. 123. Six times the product of negative five and a number

  4. 124. Ten times the product of negative four and a number

  5. 125. The difference between the product of five and a number and twice the number

  6. 126. The difference between the product of six and a number and negative two times the number

  7. 127. The difference between eight times a number and six more than three times the number

  8. 128. Eight decreased by three times the sum of a number and six

Application Exercises

The maximum heart rate, in beats per minute, that you should achieve during exercise is 220 minus your age:

220 minus a, is labeled, this algebraic expression gives maximum heart rate in terms of age, a.

The following graph shows the target heart rate ranges for four types of exercise goals. The lower and upper limits of these ranges are fractions of the maximum heart rate, 220a. Exercises 129130 are based on the information in the graph.

A horizontal bar graph titled, Target Heart Rate Ranges for Exercise Goals.
  1. 129. If your exercise goal is to improve cardiovascular conditioning, the graph shows the following range for target heart rate, H, in beats per minute:

    The lower limit of range labeled to, H = 7 tenths left parenthesis 200 minus a right parenthesis. The upper limit of range labeled to, H = 4 fifths left parenthesis 200 minus a right parenthesis.
    1. What is the lower limit of the heart rate range, in beats per minute, for a 20-year-old with this exercise goal?

    2. What is the upper limit of the heart rate range, in beats per minute, for a 20-year-old with this exercise goal?

  2. 130. If your exercise goal is to improve overall health, the graph shows the following range for target heart rate, H, in beats per minute:

    The lower limit of range labeled to, H = 1 half left parenthesis 200 minus a right parenthesis. The upper limit of range labeled to, H = 3 fifths left parenthesis 200 minus a right parenthesis.
    1. What is the lower limit of the heart rate range, in beats per minute, for a 30-year-old with this exercise goal?

    2. What is the upper limit of the heart rate range, in beats per minute, for a 30-year-old with this exercise goal?

The bar graph shows the average cost of tuition and fees at private four-year colleges in the United States.

A vertical bar graph titled, the average cost of tuition and fees at public four year U.S colleges.

Source: The College Board

Here are two formulas that model the data shown in the graph. In each formula, T represents the average cost of tuition and fees at private U.S. colleges for the school year ending x years after 2000.

Model 1 labeled to, T = 975 x + 13,547. Model 2 labeled to, T = 32 x squared + 331 x + 15,479.

Use this information to solve Exercises 131132.

  1. 131.

    1. Use each formula to find the average cost of tuition and fees at private U.S. colleges for the school year ending in 2018. By how much does each model underestimate or overestimate the actual cost shown for the school year ending in 2018?

    2. Use model 2 to project the average cost of tuition and fees at private U.S. colleges for the school year ending in 2030.

  2. 132.

    1. Use each formula to find the average cost of tuition and fees at private U.S. colleges for the school year ending in 2010. By how much does each underestimate or overestimate the actual cost shown for the school year ending in 2010?

    2. Use model 2 to project the average cost of tuition and fees at private U.S. colleges for the school year ending in 2025.

  3. 133. This month you have a total of $6000 in interest-bearing credit card debt, split between a card charging 18% annual interest and a card charging 21% annual interest. If the interest-bearing balance on the card charging 18% is x dollars, then the total interest for the month is given by the algebraic expression

    0.015x+0.0175(6000x).
    1. Simplify the algebraic expression.

    2. Use each form of the algebraic expression to determine the total interest for the month if the balance on the card charging 18% is $4400.

    3. Use the simplified form of the algebraic expression to determine the total interest for the month if the $6000 debt is split evenly between the two cards.

  4. 134. It takes you 50 minutes to get to campus. You spend t minutes walking to the bus stop and the rest of the time riding the bus. Your walking rate is 0.06 mile per minute and the bus travels at a rate of 0.5 mile per minute. The total distance walking and traveling by bus is given by the algebraic expression

    0.06t+0.5(50t).
    1. Simplify the algebraic expression.

    2. Use each form of the algebraic expression to determine the total distance that you travel if you spend 20 minutes walking to the bus stop.

    3. Use the simplified form of the algebraic expression to determine the total distance you travel if the 50 minutes is split evenly between walking and riding the bus.

  5. 135. Read the Blitzer Bonus beginning here. Use the formula

    BAC=600nw(0.6n+169)

    and replace w with your body weight. Using this formula and a calculator, compute your BAC for integers from n=1 to n=10. Round to three decimal places. According to this model, how many drinks can you consume in an hour without exceeding the legal measure of drunk driving?

Explaining the Concepts

  1. 136. What is an algebraic expression? Give an example with your explanation.

  2. 137. If n is a natural number, what does bn mean? Give an example with your explanation.

  3. 138. What does it mean when we say that a formula models real-world phenomena?

  4. 139. What is the intersection of sets A and B?

  5. 140. What is the union of sets A and B?

  6. 141. How do the whole numbers differ from the natural numbers?

  7. 142. Can a real number be both rational and irrational? Explain your answer.

  8. 143. If you are given two real numbers, explain how to determine which is the lesser.

Critical Thinking Exercises

Make Sense? In Exercises 144147, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 144. My mathematical model describes the data for tuition and fees at public four-year colleges for the past 20 years extremely well, so it will serve as an accurate prediction for the cost of public colleges in 2050.

  2. 145. A model that describes the average cost of tuition and fees at private U.S. colleges for the school year ending x years after 2000 cannot be used to estimate the cost of private education for the school year ending in 2000.

  3. 146. Regardless of what real numbers I substitute for x and y, I will always obtain zero when evaluating 2x2y2yx2.

  4. 147. Just as the commutative properties change groupings, the associative properties change order.

In Exercises 148155, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 148. Every rational number is an integer.

  2. 149. Some whole numbers are not integers.

  3. 150. Some rational numbers are not positive.

  4. 151. Irrational numbers cannot be negative.

  5. 152. The term x has no coefficient.

  6. 153. 5+3(x4)=8(x4)=8x32

  7. 154. xx=x+(x)=0

  8. 155. x0.02(x+200)=0.98x4

In Exercises 156158, insert either < or > in the shaded area between the numbers to make the statement true.

  1. 156. 2 00 1.5

  2. 157. π 00 3.5

  3. 158. 3.142 00 π2

Preview Exercises

Exercises 159161 will help you prepare for the material covered in the next section.

  1. 159. In parts (a) and (b), complete each statement.

    1. b4b3=(bbbb)(bbb)=b?

    2. b5b5=(bbbbb)(bbbbb)=b?

    3. Generalizing from parts (a) and (b), what should be done with the exponents when multiplying exponential expressions with the same base?

  2. 160. In parts (a) and (b), complete each statement.

    1. b7b3=bbbbbbbbb b=b?

    2. b8b2=bbbbbbbbbb =b?

    3. Generalizing from parts (a) and (b), what should be done with the exponents when dividing exponential expressions with the same base?

  3. 161. If 6.2 is multiplied by 103, what does this multiplication do to the decimal point in 6.2?

Objective 1: Use properties of exponents

Properties of Exponents

  1. Objective 1Use properties of exponents.

The major properties of exponents are summarized in the box that follows.

Properties of Exponents

Property Examples
The Negative-Exponent Rule

If b is any real number other than 0 and n is a natural number, then

bn=1bn.
  • 53=153=1125

  • 142=1142=42=16

The Zero-Exponent Rule

If b is any real number other than 0,

b0=1.
  • 70=1

  • (5)0=1

  • Negative 5 to the power 0 = negative 1. A note pointing to negative 5 to the power 0 reads, Only 5 is raised to the zero power.
The Product Rule

If b is a real number or algebraic expression, and m and n are integers,

bmbn=bm+n.
  • 2223=22+3=25=32

  • x3x7=x3+7=x4

When multiplying exponential expressions with the same base, add the exponents. Use this sum as the exponent of the common base.
The Power Rule

If b is a real number or algebraic expression, and m and n are integers,

(bm)n=bmn.
  • (22)3=223=26=64

  • (x3)4=x34=x12=1x12

When an exponential expression is raised to a power, multiply the exponents. Place the product of the exponents on the base and remove the parentheses.
The Quotient Rule

If b is a nonzero real number or algebraic expression, and m and n are integers,

bmbn=bmn.
  • 2824=284=24=16

  • x3x7=x37=x4=1x4

When dividing exponential expressions with the same nonzero base, subtract the exponent in the denominator from the exponent in the numerator. Use this difference as the exponent of the common base.
Products Raised to Powers

If a and b are real numbers or algebraic expressions, and n is an integer,

(ab)n=anbn.
  • (2y)4=(2)4y4=16y4

  • (2xy)3=(2)3x3y3=8x3y3

When a product is raised to a power, raise each factor to that power.
Quotients Raised to Powers

If a and b are real numbers, b0, or algebraic expressions, and n is an integer,

(ab)n=anbn.
  • (25)4=2454=16625

  • ( 3x)3=(3)3x3= 27x3

When a quotient is raised to a power, raise the numerator to that power and divide by the denominator to that power.
Objective 1: Use properties of exponents

Properties of Exponents

  1. Objective 1Use properties of exponents.

The major properties of exponents are summarized in the box that follows.

Properties of Exponents

Property Examples
The Negative-Exponent Rule

If b is any real number other than 0 and n is a natural number, then

bn=1bn.
  • 53=153=1125

  • 142=1142=42=16

The Zero-Exponent Rule

If b is any real number other than 0,

b0=1.
  • 70=1

  • (5)0=1

  • Negative 5 to the power 0 = negative 1. A note pointing to negative 5 to the power 0 reads, Only 5 is raised to the zero power.
The Product Rule

If b is a real number or algebraic expression, and m and n are integers,

bmbn=bm+n.
  • 2223=22+3=25=32

  • x3x7=x3+7=x4

When multiplying exponential expressions with the same base, add the exponents. Use this sum as the exponent of the common base.
The Power Rule

If b is a real number or algebraic expression, and m and n are integers,

(bm)n=bmn.
  • (22)3=223=26=64

  • (x3)4=x34=x12=1x12

When an exponential expression is raised to a power, multiply the exponents. Place the product of the exponents on the base and remove the parentheses.
The Quotient Rule

If b is a nonzero real number or algebraic expression, and m and n are integers,

bmbn=bmn.
  • 2824=284=24=16

  • x3x7=x37=x4=1x4

When dividing exponential expressions with the same nonzero base, subtract the exponent in the denominator from the exponent in the numerator. Use this difference as the exponent of the common base.
Products Raised to Powers

If a and b are real numbers or algebraic expressions, and n is an integer,

(ab)n=anbn.
  • (2y)4=(2)4y4=16y4

  • (2xy)3=(2)3x3y3=8x3y3

When a product is raised to a power, raise each factor to that power.
Quotients Raised to Powers

If a and b are real numbers, b0, or algebraic expressions, and n is an integer,

(ab)n=anbn.
  • (25)4=2454=16625

  • ( 3x)3=(3)3x3= 27x3

When a quotient is raised to a power, raise the numerator to that power and divide by the denominator to that power.
Objective 2: Simplify exponential expressions

Simplifying Exponential Expressions

  1. Objective 2Simplify exponential expressions.

Watch Video

Properties of exponents are used to simplify exponential expressions. An exponential expression is simplified when

Simplifying Exponential Expressions

Example

1. If necessary, remove parentheses by using

(ab)n=an bnor(ab)n=anbn.
(xy)3=x3 y3

2. If necessary, simplify powers to powers by using

(bm)n=bmn.
(x4)3=x43=x12

3. If necessary, be sure that each base appears only once by using

bmbn=bm+norbmbn=bmn.
x4x3=x4+3=x7

4. If necessary, rewrite exponential expressions with zero powers as 1 (b0=1). Furthermore, write the answer with positive exponents by using

bn=1bnor1bn=bn.
x5x8  =x58 =x3 =1x3 

The following example shows how to simplify exponential expressions. Throughout the example, assume that no variable in a denominator is equal to zero.

Example 1 Simplifying Exponential Expressions

Simplify:

  1. (3x4 y5)3

  2. (7xy4)(2x5 y6)

  3. 35x2 y45x6 y8

  4. (4x2y)3.

Solution

  1.  

    (3x4 y5)3=(3)3(x4)3(y5)3Raise each factor inside the parentheses to thethird power.=(3)3 x43y53Multiply the exponents when raising powers to powers.=27x12 y15 (3)3=(3)(3)(3)=27
  2.  

    (7xy4)(2x5 y6)=(7)(2)xx5 y4 y6 Group factors with the same base.=14x1+5 y4+6 When multiplying expressions with the  same base, add the exponents. =14x6 y10 Simplify.
  3.  

    35x2 y45x6 y8=(355)(x2x6 )(y4y8) Group factors with the same base.=7x26 y4(8)When dividing expressions with the same base, subtract the exponents.=7x4 y12 Simplify. Notice that  4(8)=4+8=12.=7y12x4Write as a fraction and move the base with thenegative exponent, x4, to the other side of thefraction bar and make the negative exponentpositive.
  4.  

    (4x2y)3=(4x2)3y3 Raise the numerator and the denominator to the 3 power.=43(x2)3y3 Raise each factor inside the parentheses to the 3  power.=43x6 y3Multiply the exponents when raising a power to a power:(x2)3=x2(3)=x6.=y343 x6Move each base with a negative exponent to the other side ofthe fraction bar and make each negative exponent positive.=y3 64x6 43=444=64

Check Point 1

  • Simplify:

    1. (2x3 y6)4

    2. (6x2 y5)(3xy3)

    3. 100x12 y220x16 y4

    4. (5xy4)2.

Objective 2: Simplify exponential expressions

Simplifying Exponential Expressions

  1. Objective 2Simplify exponential expressions.

Watch Video

Properties of exponents are used to simplify exponential expressions. An exponential expression is simplified when

Simplifying Exponential Expressions

Example

1. If necessary, remove parentheses by using

(ab)n=an bnor(ab)n=anbn.
(xy)3=x3 y3

2. If necessary, simplify powers to powers by using

(bm)n=bmn.
(x4)3=x43=x12

3. If necessary, be sure that each base appears only once by using

bmbn=bm+norbmbn=bmn.
x4x3=x4+3=x7

4. If necessary, rewrite exponential expressions with zero powers as 1 (b0=1). Furthermore, write the answer with positive exponents by using

bn=1bnor1bn=bn.
x5x8  =x58 =x3 =1x3 

The following example shows how to simplify exponential expressions. Throughout the example, assume that no variable in a denominator is equal to zero.

Example 1 Simplifying Exponential Expressions

Simplify:

  1. (3x4 y5)3

  2. (7xy4)(2x5 y6)

  3. 35x2 y45x6 y8

  4. (4x2y)3.

Solution

  1.  

    (3x4 y5)3=(3)3(x4)3(y5)3Raise each factor inside the parentheses to thethird power.=(3)3 x43y53Multiply the exponents when raising powers to powers.=27x12 y15 (3)3=(3)(3)(3)=27
  2.  

    (7xy4)(2x5 y6)=(7)(2)xx5 y4 y6 Group factors with the same base.=14x1+5 y4+6 When multiplying expressions with the  same base, add the exponents. =14x6 y10 Simplify.
  3.  

    35x2 y45x6 y8=(355)(x2x6 )(y4y8) Group factors with the same base.=7x26 y4(8)When dividing expressions with the same base, subtract the exponents.=7x4 y12 Simplify. Notice that  4(8)=4+8=12.=7y12x4Write as a fraction and move the base with thenegative exponent, x4, to the other side of thefraction bar and make the negative exponentpositive.
  4.  

    (4x2y)3=(4x2)3y3 Raise the numerator and the denominator to the 3 power.=43(x2)3y3 Raise each factor inside the parentheses to the 3  power.=43x6 y3Multiply the exponents when raising a power to a power:(x2)3=x2(3)=x6.=y343 x6Move each base with a negative exponent to the other side ofthe fraction bar and make each negative exponent positive.=y3 64x6 43=444=64

Check Point 1

  • Simplify:

    1. (2x3 y6)4

    2. (6x2 y5)(3xy3)

    3. 100x12 y220x16 y4

    4. (5xy4)2.

Objective 3: Use scientific notation

Scientific Notation

  1. Objective 3Use scientific notation.

Watch Video

Earth is a 4.5-billion-year-old ball of rock orbiting the Sun. Because a billion is 109 (see Table P.3), the age of our world, in years, can be expressed as

4.5×109.

The number 4.5×109 is written in a form called scientific notation.

Table P.3 Names of Large Numbers

102 hundred
103 thousand
106 million
109 billion
1012 trillion
1015 quadrillion
1018 quintillion
1021 sextillion
1024 septillion
1027 octillion
1030 nonillion
10100 googol
10googol googolplex

Scientific Notation

A number is written in scientific notation when it is expressed in the form

a×10n,

where the absolute value of a is greater than or equal to 1 and less than 10 (1|a|<10), and n is an integer.

It is customary to use the multiplication symbol, ×, rather than a dot, when writing a number in scientific notation.

Converting from Scientific to Decimal Notation

Here are two examples of numbers in scientific notation:

26.4×105means640,000.2.17×103 means0.00217.

Do you see that the number with the positive exponent is relatively large and the number with the negative exponent is relatively small?

We can use n, the exponent on the 10 in a×10n, to change a number in scientific notation to decimal notation. If n is positive, move the decimal point in a to the right n places. If n is negative, move the decimal point in a to the left |n| places.

Example 2 Converting from Scientific to Decimal Notation

Write each number in decimal notation:

  1. 6.2×107

  2. 6.2×107

  3. 2.019×103

  4. 2.019×103.

Solution

In each case, we use the exponent on the 10 to determine how far to move the decimal point and in which direction. In parts (a) and (b), the exponent is positive, so we move the decimal point to the right. In parts (c) and (d), the exponent is negative, so we move the decimal point to the left.

Check Point 2

  • Write each number in decimal notation:

    1. 2.6×109

    2. 3.017×106 .

Converting from Decimal to Scientific Notation

To convert from decimal notation to scientific notation, we reverse the procedure of Example 2.

Converting from Decimal to Scientific Notation

Write the number in the form a×10n.

Example 3 Converting from Decimal Notation to Scientific Notation

Write each number in scientific notation:

  1. 34,970,000,000,000

  2. 34,970,000,000,000

  3. 0.0000000000802

  4. 0.0000000000802.

Solution

The solution to write two numbers in scientific notation.

The solution to write two decimal numbers in scientific notation.

Check Point 3

  • Write each number in scientific notation:

    1. 5,210,000,000

    2. 0.00000006893.

Example 4 Expressing the U.S. Population in Scientific Notation

As of May 2020, the population of the United States was approximately 331 million. Express the population in scientific notation.

Solution

Because a million is 106, the 2020 population can be expressed as

331 times 10 to the sixth power. 331 is labeled, this factor is not between 1 and 10, so the number is not in scientific notation.

The voice balloon indicates that we need to convert 331 to scientific notation.

331 times 10 to the sixth power = left parenthesis 3.31 times 10 squared right parenthesis times 10 to the sixth power = 331 times 10 of start expression 2 + 6 end expression = 3.31 times 10 to the eighth power.

In scientific notation, the population is 3.31×108.

Check Point 4

  • Express 410×107 in scientific notation.

Computations with Scientific Notation

Properties of exponents are used to perform computations with numbers that are expressed in scientific notation.

Example 5 Computations with Scientific Notation

Perform the indicated computations, writing the answers in scientific notation:

  1. (6.1×105)(4×109)

  2. 1.8×1043×102.

Solution

  1.  

    (6.1×105)(4×109)=(6.1×4)×(105 ×109)Regroup factors.=24.4×105+(9)Add the exponents on 10 and multiply theother parts.=24.4×104Simplify.=(2.44×101)×104Convert 24.4 to scientific notation:24.4=2.44×101.=2.44×103101×104=101+(4)=103

  2.  

    1.8×104 3×102=(1.8 3)×(104102)Regroup factors.=0.6×104(2)Subtract the exponents on 10 and divide theother parts.=0.6×106Simplify:  4(2)=4+2=6.=(6×101)×106Convert 0.6 to scientific notation:0.6=6×101.=6×105101×106=101+6=105

Check Point 5

  • Perform the indicated computations, writing the answers in scientific notation:

    1. (7.1×105)(5×107)

    2. 1.2×1063×103.

Applications: Putting Numbers in Perspective

Due to tax cuts and spending increases, the United States began accumulating large deficits in the 1980s. To finance the deficit, the government had borrowed $25.5 trillion as of May 2020. The graph in Figure P.10 shows the national debt increasing over time. The amount shown for 2020 is midyear; the big increase in the national debt in the early part of the year was due in part to the economic impact of the COVID-19 pandemic.

Figure P.10

A vertical bar graph titled, The National Debt.

Source: Office of Management and Budget

Figure P.10 Full Alternative Text

Example 6 shows how we can use scientific notation to comprehend the meaning of a number such as 25.5 trillion.

Example 6 The National Debt

As of May 2020, the national debt was $25.5 trillion, or 25.5×1012 dollars. At that time, the U.S. population was approximately 331,000,000 (331 million), or 3.31×108. If the national debt was evenly divided among every individual in the United States, how much would each citizen have to pay?

Solution

The amount each citizen must pay is the total debt, 25.5×1012 dollars, divided by the number of citizens, 3.31×108.

25.5×10123.31×108=(25.53.31)×(1012108)7.70×10128=7.70×104=77,000

Every U.S. citizen would have to pay approximately $77,000 to the federal government to pay off the national debt. At the end of 2019, the median yearly income of full-time workers in the United States was $48,672 (Source: Bureau of Labor Statistics); a worker with this income would have to pay more than 1.5 years’ salary to pay off the national debt.

Check Point 6

  • At the end of 2010, the national debt was $13.5 trillion. At that time, the U.S. population was approximately 309 million. Write 13.5 trillion and 309 million in scientific notation. Then determine how much each citizen would have had to pay if the national debt in 2010 had been evenly divided among every individual in the 2010 population. Round to the nearest thousand dollars. Compare this amount to the $77,000 per person in 2020 from Example 6. How much did each individual’s share of the national debt increase between 2010 and 2020?

An Application: Black Holes in Space

The concept of a black hole, a region in space where matter appears to vanish, intrigues scientists and nonscientists alike. Scientists theorize that when massive stars run out of nuclear fuel, they begin to collapse under the force of their own gravity. As the star collapses, its density increases. In turn, the force of gravity increases so tremendously that even light cannot escape from the star. Consequently, it appears black.

A mathematical model, called the Schwarzchild formula, describes the critical value to which the radius of a massive body must be reduced for it to become a black hole. This model forms the basis of our next example.

Example 7 An Application of Scientific Notation

Use the Schwarzchild formula

Rs=2GMc2

where

Rs=Radius of the star,in meters,that would cause it to become a black holeM=Mass of the star,in kilogramsG=A constant,called the gravitational constant=6.7×1011m3kgs2c=Speed of light=3×108 meters per second

to determine to what length the radius of the Sun must be reduced for it to become a black hole. The Sun’s mass is approximately 2×1030 kilograms.

Solution

 Rs=2GMc2Use the given model.=2×6.7×1011×2×1030(3×108)2Substitute the given values:G=6.7×1011,M=2×1030, andc=3×108.=(2×6.7×2)×(1011×1030)(3×108)2Rearrange factors in the numerator.=26.8×1011+3032×(108)2Add exponents in the numerator. Raise eachfactor in the denominator to the power.=26.8×10199×1016Multiply powers to powers:(108)2=108.2=1016.=26.89×101916When dividing expressions with the samebase,subtract the exponents. 2.978×103Simplify.=2978

Although the Sun is not massive enough to become a black hole (its radius is approximately 700,000 kilometers), the Schwarzchild model theoretically indicates that if the Sun’s radius were reduced to approximately 2978 meters, that is, about 1235,000 its present size, it would become a black hole.

Check Point 7

  • The speed of blood, S, in centimeters per second, located r centimeters from the central axis of an artery is modeled by

    S=(1.76×105)[(1.44×102)r2].

    Find the speed of blood at the central axis of this artery.

Objective 3: Use scientific notation

Scientific Notation

  1. Objective 3Use scientific notation.

Watch Video

Earth is a 4.5-billion-year-old ball of rock orbiting the Sun. Because a billion is 109 (see Table P.3), the age of our world, in years, can be expressed as

4.5×109.

The number 4.5×109 is written in a form called scientific notation.

Table P.3 Names of Large Numbers

102 hundred
103 thousand
106 million
109 billion
1012 trillion
1015 quadrillion
1018 quintillion
1021 sextillion
1024 septillion
1027 octillion
1030 nonillion
10100 googol
10googol googolplex

Scientific Notation

A number is written in scientific notation when it is expressed in the form

a×10n,

where the absolute value of a is greater than or equal to 1 and less than 10 (1|a|<10), and n is an integer.

It is customary to use the multiplication symbol, ×, rather than a dot, when writing a number in scientific notation.

Converting from Scientific to Decimal Notation

Here are two examples of numbers in scientific notation:

26.4×105means640,000.2.17×103 means0.00217.

Do you see that the number with the positive exponent is relatively large and the number with the negative exponent is relatively small?

We can use n, the exponent on the 10 in a×10n, to change a number in scientific notation to decimal notation. If n is positive, move the decimal point in a to the right n places. If n is negative, move the decimal point in a to the left |n| places.

Example 2 Converting from Scientific to Decimal Notation

Write each number in decimal notation:

  1. 6.2×107

  2. 6.2×107

  3. 2.019×103

  4. 2.019×103.

Solution

In each case, we use the exponent on the 10 to determine how far to move the decimal point and in which direction. In parts (a) and (b), the exponent is positive, so we move the decimal point to the right. In parts (c) and (d), the exponent is negative, so we move the decimal point to the left.

Check Point 2

  • Write each number in decimal notation:

    1. 2.6×109

    2. 3.017×106 .

Converting from Decimal to Scientific Notation

To convert from decimal notation to scientific notation, we reverse the procedure of Example 2.

Converting from Decimal to Scientific Notation

Write the number in the form a×10n.

Example 3 Converting from Decimal Notation to Scientific Notation

Write each number in scientific notation:

  1. 34,970,000,000,000

  2. 34,970,000,000,000

  3. 0.0000000000802

  4. 0.0000000000802.

Solution

The solution to write two numbers in scientific notation.

The solution to write two decimal numbers in scientific notation.

Check Point 3

  • Write each number in scientific notation:

    1. 5,210,000,000

    2. 0.00000006893.

Example 4 Expressing the U.S. Population in Scientific Notation

As of May 2020, the population of the United States was approximately 331 million. Express the population in scientific notation.

Solution

Because a million is 106, the 2020 population can be expressed as

331 times 10 to the sixth power. 331 is labeled, this factor is not between 1 and 10, so the number is not in scientific notation.

The voice balloon indicates that we need to convert 331 to scientific notation.

331 times 10 to the sixth power = left parenthesis 3.31 times 10 squared right parenthesis times 10 to the sixth power = 331 times 10 of start expression 2 + 6 end expression = 3.31 times 10 to the eighth power.

In scientific notation, the population is 3.31×108.

Check Point 4

  • Express 410×107 in scientific notation.

Computations with Scientific Notation

Properties of exponents are used to perform computations with numbers that are expressed in scientific notation.

Example 5 Computations with Scientific Notation

Perform the indicated computations, writing the answers in scientific notation:

  1. (6.1×105)(4×109)

  2. 1.8×1043×102.

Solution

  1.  

    (6.1×105)(4×109)=(6.1×4)×(105 ×109)Regroup factors.=24.4×105+(9)Add the exponents on 10 and multiply theother parts.=24.4×104Simplify.=(2.44×101)×104Convert 24.4 to scientific notation:24.4=2.44×101.=2.44×103101×104=101+(4)=103

  2.  

    1.8×104 3×102=(1.8 3)×(104102)Regroup factors.=0.6×104(2)Subtract the exponents on 10 and divide theother parts.=0.6×106Simplify:  4(2)=4+2=6.=(6×101)×106Convert 0.6 to scientific notation:0.6=6×101.=6×105101×106=101+6=105

Check Point 5

  • Perform the indicated computations, writing the answers in scientific notation:

    1. (7.1×105)(5×107)

    2. 1.2×1063×103.

Applications: Putting Numbers in Perspective

Due to tax cuts and spending increases, the United States began accumulating large deficits in the 1980s. To finance the deficit, the government had borrowed $25.5 trillion as of May 2020. The graph in Figure P.10 shows the national debt increasing over time. The amount shown for 2020 is midyear; the big increase in the national debt in the early part of the year was due in part to the economic impact of the COVID-19 pandemic.

Figure P.10

A vertical bar graph titled, The National Debt.

Source: Office of Management and Budget

Figure P.10 Full Alternative Text

Example 6 shows how we can use scientific notation to comprehend the meaning of a number such as 25.5 trillion.

Example 6 The National Debt

As of May 2020, the national debt was $25.5 trillion, or 25.5×1012 dollars. At that time, the U.S. population was approximately 331,000,000 (331 million), or 3.31×108. If the national debt was evenly divided among every individual in the United States, how much would each citizen have to pay?

Solution

The amount each citizen must pay is the total debt, 25.5×1012 dollars, divided by the number of citizens, 3.31×108.

25.5×10123.31×108=(25.53.31)×(1012108)7.70×10128=7.70×104=77,000

Every U.S. citizen would have to pay approximately $77,000 to the federal government to pay off the national debt. At the end of 2019, the median yearly income of full-time workers in the United States was $48,672 (Source: Bureau of Labor Statistics); a worker with this income would have to pay more than 1.5 years’ salary to pay off the national debt.

Check Point 6

  • At the end of 2010, the national debt was $13.5 trillion. At that time, the U.S. population was approximately 309 million. Write 13.5 trillion and 309 million in scientific notation. Then determine how much each citizen would have had to pay if the national debt in 2010 had been evenly divided among every individual in the 2010 population. Round to the nearest thousand dollars. Compare this amount to the $77,000 per person in 2020 from Example 6. How much did each individual’s share of the national debt increase between 2010 and 2020?

An Application: Black Holes in Space

The concept of a black hole, a region in space where matter appears to vanish, intrigues scientists and nonscientists alike. Scientists theorize that when massive stars run out of nuclear fuel, they begin to collapse under the force of their own gravity. As the star collapses, its density increases. In turn, the force of gravity increases so tremendously that even light cannot escape from the star. Consequently, it appears black.

A mathematical model, called the Schwarzchild formula, describes the critical value to which the radius of a massive body must be reduced for it to become a black hole. This model forms the basis of our next example.

Example 7 An Application of Scientific Notation

Use the Schwarzchild formula

Rs=2GMc2

where

Rs=Radius of the star,in meters,that would cause it to become a black holeM=Mass of the star,in kilogramsG=A constant,called the gravitational constant=6.7×1011m3kgs2c=Speed of light=3×108 meters per second

to determine to what length the radius of the Sun must be reduced for it to become a black hole. The Sun’s mass is approximately 2×1030 kilograms.

Solution

 Rs=2GMc2Use the given model.=2×6.7×1011×2×1030(3×108)2Substitute the given values:G=6.7×1011,M=2×1030, andc=3×108.=(2×6.7×2)×(1011×1030)(3×108)2Rearrange factors in the numerator.=26.8×1011+3032×(108)2Add exponents in the numerator. Raise eachfactor in the denominator to the power.=26.8×10199×1016Multiply powers to powers:(108)2=108.2=1016.=26.89×101916When dividing expressions with the samebase,subtract the exponents. 2.978×103Simplify.=2978

Although the Sun is not massive enough to become a black hole (its radius is approximately 700,000 kilometers), the Schwarzchild model theoretically indicates that if the Sun’s radius were reduced to approximately 2978 meters, that is, about 1235,000 its present size, it would become a black hole.

Check Point 7

  • The speed of blood, S, in centimeters per second, located r centimeters from the central axis of an artery is modeled by

    S=(1.76×105)[(1.44×102)r2].

    Find the speed of blood at the central axis of this artery.

Objective 3: Use scientific notation

Scientific Notation

  1. Objective 3Use scientific notation.

Watch Video

Earth is a 4.5-billion-year-old ball of rock orbiting the Sun. Because a billion is 109 (see Table P.3), the age of our world, in years, can be expressed as

4.5×109.

The number 4.5×109 is written in a form called scientific notation.

Table P.3 Names of Large Numbers

102 hundred
103 thousand
106 million
109 billion
1012 trillion
1015 quadrillion
1018 quintillion
1021 sextillion
1024 septillion
1027 octillion
1030 nonillion
10100 googol
10googol googolplex

Scientific Notation

A number is written in scientific notation when it is expressed in the form

a×10n,

where the absolute value of a is greater than or equal to 1 and less than 10 (1|a|<10), and n is an integer.

It is customary to use the multiplication symbol, ×, rather than a dot, when writing a number in scientific notation.

Converting from Scientific to Decimal Notation

Here are two examples of numbers in scientific notation:

26.4×105means640,000.2.17×103 means0.00217.

Do you see that the number with the positive exponent is relatively large and the number with the negative exponent is relatively small?

We can use n, the exponent on the 10 in a×10n, to change a number in scientific notation to decimal notation. If n is positive, move the decimal point in a to the right n places. If n is negative, move the decimal point in a to the left |n| places.

Example 2 Converting from Scientific to Decimal Notation

Write each number in decimal notation:

  1. 6.2×107

  2. 6.2×107

  3. 2.019×103

  4. 2.019×103.

Solution

In each case, we use the exponent on the 10 to determine how far to move the decimal point and in which direction. In parts (a) and (b), the exponent is positive, so we move the decimal point to the right. In parts (c) and (d), the exponent is negative, so we move the decimal point to the left.

Check Point 2

  • Write each number in decimal notation:

    1. 2.6×109

    2. 3.017×106 .

Converting from Decimal to Scientific Notation

To convert from decimal notation to scientific notation, we reverse the procedure of Example 2.

Converting from Decimal to Scientific Notation

Write the number in the form a×10n.

Example 3 Converting from Decimal Notation to Scientific Notation

Write each number in scientific notation:

  1. 34,970,000,000,000

  2. 34,970,000,000,000

  3. 0.0000000000802

  4. 0.0000000000802.

Solution

The solution to write two numbers in scientific notation.

The solution to write two decimal numbers in scientific notation.

Check Point 3

  • Write each number in scientific notation:

    1. 5,210,000,000

    2. 0.00000006893.

Example 4 Expressing the U.S. Population in Scientific Notation

As of May 2020, the population of the United States was approximately 331 million. Express the population in scientific notation.

Solution

Because a million is 106, the 2020 population can be expressed as

331 times 10 to the sixth power. 331 is labeled, this factor is not between 1 and 10, so the number is not in scientific notation.

The voice balloon indicates that we need to convert 331 to scientific notation.

331 times 10 to the sixth power = left parenthesis 3.31 times 10 squared right parenthesis times 10 to the sixth power = 331 times 10 of start expression 2 + 6 end expression = 3.31 times 10 to the eighth power.

In scientific notation, the population is 3.31×108.

Check Point 4

  • Express 410×107 in scientific notation.

Computations with Scientific Notation

Properties of exponents are used to perform computations with numbers that are expressed in scientific notation.

Example 5 Computations with Scientific Notation

Perform the indicated computations, writing the answers in scientific notation:

  1. (6.1×105)(4×109)

  2. 1.8×1043×102.

Solution

  1.  

    (6.1×105)(4×109)=(6.1×4)×(105 ×109)Regroup factors.=24.4×105+(9)Add the exponents on 10 and multiply theother parts.=24.4×104Simplify.=(2.44×101)×104Convert 24.4 to scientific notation:24.4=2.44×101.=2.44×103101×104=101+(4)=103

  2.  

    1.8×104 3×102=(1.8 3)×(104102)Regroup factors.=0.6×104(2)Subtract the exponents on 10 and divide theother parts.=0.6×106Simplify:  4(2)=4+2=6.=(6×101)×106Convert 0.6 to scientific notation:0.6=6×101.=6×105101×106=101+6=105

Check Point 5

  • Perform the indicated computations, writing the answers in scientific notation:

    1. (7.1×105)(5×107)

    2. 1.2×1063×103.

Applications: Putting Numbers in Perspective

Due to tax cuts and spending increases, the United States began accumulating large deficits in the 1980s. To finance the deficit, the government had borrowed $25.5 trillion as of May 2020. The graph in Figure P.10 shows the national debt increasing over time. The amount shown for 2020 is midyear; the big increase in the national debt in the early part of the year was due in part to the economic impact of the COVID-19 pandemic.

Figure P.10

A vertical bar graph titled, The National Debt.

Source: Office of Management and Budget

Figure P.10 Full Alternative Text

Example 6 shows how we can use scientific notation to comprehend the meaning of a number such as 25.5 trillion.

Example 6 The National Debt

As of May 2020, the national debt was $25.5 trillion, or 25.5×1012 dollars. At that time, the U.S. population was approximately 331,000,000 (331 million), or 3.31×108. If the national debt was evenly divided among every individual in the United States, how much would each citizen have to pay?

Solution

The amount each citizen must pay is the total debt, 25.5×1012 dollars, divided by the number of citizens, 3.31×108.

25.5×10123.31×108=(25.53.31)×(1012108)7.70×10128=7.70×104=77,000

Every U.S. citizen would have to pay approximately $77,000 to the federal government to pay off the national debt. At the end of 2019, the median yearly income of full-time workers in the United States was $48,672 (Source: Bureau of Labor Statistics); a worker with this income would have to pay more than 1.5 years’ salary to pay off the national debt.

Check Point 6

  • At the end of 2010, the national debt was $13.5 trillion. At that time, the U.S. population was approximately 309 million. Write 13.5 trillion and 309 million in scientific notation. Then determine how much each citizen would have had to pay if the national debt in 2010 had been evenly divided among every individual in the 2010 population. Round to the nearest thousand dollars. Compare this amount to the $77,000 per person in 2020 from Example 6. How much did each individual’s share of the national debt increase between 2010 and 2020?

An Application: Black Holes in Space

The concept of a black hole, a region in space where matter appears to vanish, intrigues scientists and nonscientists alike. Scientists theorize that when massive stars run out of nuclear fuel, they begin to collapse under the force of their own gravity. As the star collapses, its density increases. In turn, the force of gravity increases so tremendously that even light cannot escape from the star. Consequently, it appears black.

A mathematical model, called the Schwarzchild formula, describes the critical value to which the radius of a massive body must be reduced for it to become a black hole. This model forms the basis of our next example.

Example 7 An Application of Scientific Notation

Use the Schwarzchild formula

Rs=2GMc2

where

Rs=Radius of the star,in meters,that would cause it to become a black holeM=Mass of the star,in kilogramsG=A constant,called the gravitational constant=6.7×1011m3kgs2c=Speed of light=3×108 meters per second

to determine to what length the radius of the Sun must be reduced for it to become a black hole. The Sun’s mass is approximately 2×1030 kilograms.

Solution

 Rs=2GMc2Use the given model.=2×6.7×1011×2×1030(3×108)2Substitute the given values:G=6.7×1011,M=2×1030, andc=3×108.=(2×6.7×2)×(1011×1030)(3×108)2Rearrange factors in the numerator.=26.8×1011+3032×(108)2Add exponents in the numerator. Raise eachfactor in the denominator to the power.=26.8×10199×1016Multiply powers to powers:(108)2=108.2=1016.=26.89×101916When dividing expressions with the samebase,subtract the exponents. 2.978×103Simplify.=2978

Although the Sun is not massive enough to become a black hole (its radius is approximately 700,000 kilometers), the Schwarzchild model theoretically indicates that if the Sun’s radius were reduced to approximately 2978 meters, that is, about 1235,000 its present size, it would become a black hole.

Check Point 7

  • The speed of blood, S, in centimeters per second, located r centimeters from the central axis of an artery is modeled by

    S=(1.76×105)[(1.44×102)r2].

    Find the speed of blood at the central axis of this artery.

Objective 3: Use scientific notation

Scientific Notation

  1. Objective 3Use scientific notation.

Watch Video

Earth is a 4.5-billion-year-old ball of rock orbiting the Sun. Because a billion is 109 (see Table P.3), the age of our world, in years, can be expressed as

4.5×109.

The number 4.5×109 is written in a form called scientific notation.

Table P.3 Names of Large Numbers

102 hundred
103 thousand
106 million
109 billion
1012 trillion
1015 quadrillion
1018 quintillion
1021 sextillion
1024 septillion
1027 octillion
1030 nonillion
10100 googol
10googol googolplex

Scientific Notation

A number is written in scientific notation when it is expressed in the form

a×10n,

where the absolute value of a is greater than or equal to 1 and less than 10 (1|a|<10), and n is an integer.

It is customary to use the multiplication symbol, ×, rather than a dot, when writing a number in scientific notation.

Converting from Scientific to Decimal Notation

Here are two examples of numbers in scientific notation:

26.4×105means640,000.2.17×103 means0.00217.

Do you see that the number with the positive exponent is relatively large and the number with the negative exponent is relatively small?

We can use n, the exponent on the 10 in a×10n, to change a number in scientific notation to decimal notation. If n is positive, move the decimal point in a to the right n places. If n is negative, move the decimal point in a to the left |n| places.

Example 2 Converting from Scientific to Decimal Notation

Write each number in decimal notation:

  1. 6.2×107

  2. 6.2×107

  3. 2.019×103

  4. 2.019×103.

Solution

In each case, we use the exponent on the 10 to determine how far to move the decimal point and in which direction. In parts (a) and (b), the exponent is positive, so we move the decimal point to the right. In parts (c) and (d), the exponent is negative, so we move the decimal point to the left.

Check Point 2

  • Write each number in decimal notation:

    1. 2.6×109

    2. 3.017×106 .

Converting from Decimal to Scientific Notation

To convert from decimal notation to scientific notation, we reverse the procedure of Example 2.

Converting from Decimal to Scientific Notation

Write the number in the form a×10n.

Example 3 Converting from Decimal Notation to Scientific Notation

Write each number in scientific notation:

  1. 34,970,000,000,000

  2. 34,970,000,000,000

  3. 0.0000000000802

  4. 0.0000000000802.

Solution

The solution to write two numbers in scientific notation.

The solution to write two decimal numbers in scientific notation.

Check Point 3

  • Write each number in scientific notation:

    1. 5,210,000,000

    2. 0.00000006893.

Example 4 Expressing the U.S. Population in Scientific Notation

As of May 2020, the population of the United States was approximately 331 million. Express the population in scientific notation.

Solution

Because a million is 106, the 2020 population can be expressed as

331 times 10 to the sixth power. 331 is labeled, this factor is not between 1 and 10, so the number is not in scientific notation.

The voice balloon indicates that we need to convert 331 to scientific notation.

331 times 10 to the sixth power = left parenthesis 3.31 times 10 squared right parenthesis times 10 to the sixth power = 331 times 10 of start expression 2 + 6 end expression = 3.31 times 10 to the eighth power.

In scientific notation, the population is 3.31×108.

Check Point 4

  • Express 410×107 in scientific notation.

Computations with Scientific Notation

Properties of exponents are used to perform computations with numbers that are expressed in scientific notation.

Example 5 Computations with Scientific Notation

Perform the indicated computations, writing the answers in scientific notation:

  1. (6.1×105)(4×109)

  2. 1.8×1043×102.

Solution

  1.  

    (6.1×105)(4×109)=(6.1×4)×(105 ×109)Regroup factors.=24.4×105+(9)Add the exponents on 10 and multiply theother parts.=24.4×104Simplify.=(2.44×101)×104Convert 24.4 to scientific notation:24.4=2.44×101.=2.44×103101×104=101+(4)=103

  2.  

    1.8×104 3×102=(1.8 3)×(104102)Regroup factors.=0.6×104(2)Subtract the exponents on 10 and divide theother parts.=0.6×106Simplify:  4(2)=4+2=6.=(6×101)×106Convert 0.6 to scientific notation:0.6=6×101.=6×105101×106=101+6=105

Check Point 5

  • Perform the indicated computations, writing the answers in scientific notation:

    1. (7.1×105)(5×107)

    2. 1.2×1063×103.

Applications: Putting Numbers in Perspective

Due to tax cuts and spending increases, the United States began accumulating large deficits in the 1980s. To finance the deficit, the government had borrowed $25.5 trillion as of May 2020. The graph in Figure P.10 shows the national debt increasing over time. The amount shown for 2020 is midyear; the big increase in the national debt in the early part of the year was due in part to the economic impact of the COVID-19 pandemic.

Figure P.10

A vertical bar graph titled, The National Debt.

Source: Office of Management and Budget

Figure P.10 Full Alternative Text

Example 6 shows how we can use scientific notation to comprehend the meaning of a number such as 25.5 trillion.

Example 6 The National Debt

As of May 2020, the national debt was $25.5 trillion, or 25.5×1012 dollars. At that time, the U.S. population was approximately 331,000,000 (331 million), or 3.31×108. If the national debt was evenly divided among every individual in the United States, how much would each citizen have to pay?

Solution

The amount each citizen must pay is the total debt, 25.5×1012 dollars, divided by the number of citizens, 3.31×108.

25.5×10123.31×108=(25.53.31)×(1012108)7.70×10128=7.70×104=77,000

Every U.S. citizen would have to pay approximately $77,000 to the federal government to pay off the national debt. At the end of 2019, the median yearly income of full-time workers in the United States was $48,672 (Source: Bureau of Labor Statistics); a worker with this income would have to pay more than 1.5 years’ salary to pay off the national debt.

Check Point 6

  • At the end of 2010, the national debt was $13.5 trillion. At that time, the U.S. population was approximately 309 million. Write 13.5 trillion and 309 million in scientific notation. Then determine how much each citizen would have had to pay if the national debt in 2010 had been evenly divided among every individual in the 2010 population. Round to the nearest thousand dollars. Compare this amount to the $77,000 per person in 2020 from Example 6. How much did each individual’s share of the national debt increase between 2010 and 2020?

An Application: Black Holes in Space

The concept of a black hole, a region in space where matter appears to vanish, intrigues scientists and nonscientists alike. Scientists theorize that when massive stars run out of nuclear fuel, they begin to collapse under the force of their own gravity. As the star collapses, its density increases. In turn, the force of gravity increases so tremendously that even light cannot escape from the star. Consequently, it appears black.

A mathematical model, called the Schwarzchild formula, describes the critical value to which the radius of a massive body must be reduced for it to become a black hole. This model forms the basis of our next example.

Example 7 An Application of Scientific Notation

Use the Schwarzchild formula

Rs=2GMc2

where

Rs=Radius of the star,in meters,that would cause it to become a black holeM=Mass of the star,in kilogramsG=A constant,called the gravitational constant=6.7×1011m3kgs2c=Speed of light=3×108 meters per second

to determine to what length the radius of the Sun must be reduced for it to become a black hole. The Sun’s mass is approximately 2×1030 kilograms.

Solution

 Rs=2GMc2Use the given model.=2×6.7×1011×2×1030(3×108)2Substitute the given values:G=6.7×1011,M=2×1030, andc=3×108.=(2×6.7×2)×(1011×1030)(3×108)2Rearrange factors in the numerator.=26.8×1011+3032×(108)2Add exponents in the numerator. Raise eachfactor in the denominator to the power.=26.8×10199×1016Multiply powers to powers:(108)2=108.2=1016.=26.89×101916When dividing expressions with the samebase,subtract the exponents. 2.978×103Simplify.=2978

Although the Sun is not massive enough to become a black hole (its radius is approximately 700,000 kilometers), the Schwarzchild model theoretically indicates that if the Sun’s radius were reduced to approximately 2978 meters, that is, about 1235,000 its present size, it would become a black hole.

Check Point 7

  • The speed of blood, S, in centimeters per second, located r centimeters from the central axis of an artery is modeled by

    S=(1.76×105)[(1.44×102)r2].

    Find the speed of blood at the central axis of this artery.

Objective 3: Use scientific notation

Scientific Notation

  1. Objective 3Use scientific notation.

Watch Video

Earth is a 4.5-billion-year-old ball of rock orbiting the Sun. Because a billion is 109 (see Table P.3), the age of our world, in years, can be expressed as

4.5×109.

The number 4.5×109 is written in a form called scientific notation.

Table P.3 Names of Large Numbers

102 hundred
103 thousand
106 million
109 billion
1012 trillion
1015 quadrillion
1018 quintillion
1021 sextillion
1024 septillion
1027 octillion
1030 nonillion
10100 googol
10googol googolplex

Scientific Notation

A number is written in scientific notation when it is expressed in the form

a×10n,

where the absolute value of a is greater than or equal to 1 and less than 10 (1|a|<10), and n is an integer.

It is customary to use the multiplication symbol, ×, rather than a dot, when writing a number in scientific notation.

Converting from Scientific to Decimal Notation

Here are two examples of numbers in scientific notation:

26.4×105means640,000.2.17×103 means0.00217.

Do you see that the number with the positive exponent is relatively large and the number with the negative exponent is relatively small?

We can use n, the exponent on the 10 in a×10n, to change a number in scientific notation to decimal notation. If n is positive, move the decimal point in a to the right n places. If n is negative, move the decimal point in a to the left |n| places.

Example 2 Converting from Scientific to Decimal Notation

Write each number in decimal notation:

  1. 6.2×107

  2. 6.2×107

  3. 2.019×103

  4. 2.019×103.

Solution

In each case, we use the exponent on the 10 to determine how far to move the decimal point and in which direction. In parts (a) and (b), the exponent is positive, so we move the decimal point to the right. In parts (c) and (d), the exponent is negative, so we move the decimal point to the left.

Check Point 2

  • Write each number in decimal notation:

    1. 2.6×109

    2. 3.017×106 .

Converting from Decimal to Scientific Notation

To convert from decimal notation to scientific notation, we reverse the procedure of Example 2.

Converting from Decimal to Scientific Notation

Write the number in the form a×10n.

Example 3 Converting from Decimal Notation to Scientific Notation

Write each number in scientific notation:

  1. 34,970,000,000,000

  2. 34,970,000,000,000

  3. 0.0000000000802

  4. 0.0000000000802.

Solution

The solution to write two numbers in scientific notation.

The solution to write two decimal numbers in scientific notation.

Check Point 3

  • Write each number in scientific notation:

    1. 5,210,000,000

    2. 0.00000006893.

Example 4 Expressing the U.S. Population in Scientific Notation

As of May 2020, the population of the United States was approximately 331 million. Express the population in scientific notation.

Solution

Because a million is 106, the 2020 population can be expressed as

331 times 10 to the sixth power. 331 is labeled, this factor is not between 1 and 10, so the number is not in scientific notation.

The voice balloon indicates that we need to convert 331 to scientific notation.

331 times 10 to the sixth power = left parenthesis 3.31 times 10 squared right parenthesis times 10 to the sixth power = 331 times 10 of start expression 2 + 6 end expression = 3.31 times 10 to the eighth power.

In scientific notation, the population is 3.31×108.

Check Point 4

  • Express 410×107 in scientific notation.

Computations with Scientific Notation

Properties of exponents are used to perform computations with numbers that are expressed in scientific notation.

Example 5 Computations with Scientific Notation

Perform the indicated computations, writing the answers in scientific notation:

  1. (6.1×105)(4×109)

  2. 1.8×1043×102.

Solution

  1.  

    (6.1×105)(4×109)=(6.1×4)×(105 ×109)Regroup factors.=24.4×105+(9)Add the exponents on 10 and multiply theother parts.=24.4×104Simplify.=(2.44×101)×104Convert 24.4 to scientific notation:24.4=2.44×101.=2.44×103101×104=101+(4)=103

  2.  

    1.8×104 3×102=(1.8 3)×(104102)Regroup factors.=0.6×104(2)Subtract the exponents on 10 and divide theother parts.=0.6×106Simplify:  4(2)=4+2=6.=(6×101)×106Convert 0.6 to scientific notation:0.6=6×101.=6×105101×106=101+6=105

Check Point 5

  • Perform the indicated computations, writing the answers in scientific notation:

    1. (7.1×105)(5×107)

    2. 1.2×1063×103.

Applications: Putting Numbers in Perspective

Due to tax cuts and spending increases, the United States began accumulating large deficits in the 1980s. To finance the deficit, the government had borrowed $25.5 trillion as of May 2020. The graph in Figure P.10 shows the national debt increasing over time. The amount shown for 2020 is midyear; the big increase in the national debt in the early part of the year was due in part to the economic impact of the COVID-19 pandemic.

Figure P.10

A vertical bar graph titled, The National Debt.

Source: Office of Management and Budget

Figure P.10 Full Alternative Text

Example 6 shows how we can use scientific notation to comprehend the meaning of a number such as 25.5 trillion.

Example 6 The National Debt

As of May 2020, the national debt was $25.5 trillion, or 25.5×1012 dollars. At that time, the U.S. population was approximately 331,000,000 (331 million), or 3.31×108. If the national debt was evenly divided among every individual in the United States, how much would each citizen have to pay?

Solution

The amount each citizen must pay is the total debt, 25.5×1012 dollars, divided by the number of citizens, 3.31×108.

25.5×10123.31×108=(25.53.31)×(1012108)7.70×10128=7.70×104=77,000

Every U.S. citizen would have to pay approximately $77,000 to the federal government to pay off the national debt. At the end of 2019, the median yearly income of full-time workers in the United States was $48,672 (Source: Bureau of Labor Statistics); a worker with this income would have to pay more than 1.5 years’ salary to pay off the national debt.

Check Point 6

  • At the end of 2010, the national debt was $13.5 trillion. At that time, the U.S. population was approximately 309 million. Write 13.5 trillion and 309 million in scientific notation. Then determine how much each citizen would have had to pay if the national debt in 2010 had been evenly divided among every individual in the 2010 population. Round to the nearest thousand dollars. Compare this amount to the $77,000 per person in 2020 from Example 6. How much did each individual’s share of the national debt increase between 2010 and 2020?

An Application: Black Holes in Space

The concept of a black hole, a region in space where matter appears to vanish, intrigues scientists and nonscientists alike. Scientists theorize that when massive stars run out of nuclear fuel, they begin to collapse under the force of their own gravity. As the star collapses, its density increases. In turn, the force of gravity increases so tremendously that even light cannot escape from the star. Consequently, it appears black.

A mathematical model, called the Schwarzchild formula, describes the critical value to which the radius of a massive body must be reduced for it to become a black hole. This model forms the basis of our next example.

Example 7 An Application of Scientific Notation

Use the Schwarzchild formula

Rs=2GMc2

where

Rs=Radius of the star,in meters,that would cause it to become a black holeM=Mass of the star,in kilogramsG=A constant,called the gravitational constant=6.7×1011m3kgs2c=Speed of light=3×108 meters per second

to determine to what length the radius of the Sun must be reduced for it to become a black hole. The Sun’s mass is approximately 2×1030 kilograms.

Solution

 Rs=2GMc2Use the given model.=2×6.7×1011×2×1030(3×108)2Substitute the given values:G=6.7×1011,M=2×1030, andc=3×108.=(2×6.7×2)×(1011×1030)(3×108)2Rearrange factors in the numerator.=26.8×1011+3032×(108)2Add exponents in the numerator. Raise eachfactor in the denominator to the power.=26.8×10199×1016Multiply powers to powers:(108)2=108.2=1016.=26.89×101916When dividing expressions with the samebase,subtract the exponents. 2.978×103Simplify.=2978

Although the Sun is not massive enough to become a black hole (its radius is approximately 700,000 kilometers), the Schwarzchild model theoretically indicates that if the Sun’s radius were reduced to approximately 2978 meters, that is, about 1235,000 its present size, it would become a black hole.

Check Point 7

  • The speed of blood, S, in centimeters per second, located r centimeters from the central axis of an artery is modeled by

    S=(1.76×105)[(1.44×102)r2].

    Find the speed of blood at the central axis of this artery.

P.2: Concept and Vocabulary Check

P.2 Concept and Vocabulary Check

Fill in each blank so that the resulting statement is true.

  1. C1. The product rule for exponents states that bmbn= _______. When multiplying exponential expressions with the same base, ______ the exponents.

  2. C2. The quotient rule for exponents states that bmbn= _______, b0. When dividing exponential expressions with the same nonzero base, ______ the exponents.

  3. C3. If b0, then b0= _______.

  4. C4. The negative-exponent rule states that bn= _______, b0.

  5. C5. True or false: 52=52 _______

  6. C6. Negative exponents in denominators can be evaluated using 1bn= ______, b0.

  7. C7. True or false: 182=82 _______

  8. C8. A positive number is written in scientific notation when it is expressed in the form a×10n, where a is ________________________ and n is a/an ______.

  9. C9. True or false: 4×103 is written in scientific notation. ______

  10. C10. True or false: 40×102 is written in scientific notation. ______

P.2: Exercise Set

P.2 Exercise Set

Practice Exercises

Evaluate each exponential expression in Exercises 122.

  1. 1. 522

  2. 2. 622

  3. 3. (2)6

  4. 4. (2)4

  5. 5. 26

  6. 6. 24

  7. 7. (3)0

  8. 8. (9)0

  9. 9. 30

  10. 10. 90

  11. 11. 43

  12. 12. 26

  13. 13. 2223

  14. 14. 3332

  15. 15. (22)3

  16. 16. (33)2

  17. 17. 2824

  18. 18. 3834

  19. 19. 333

  20. 20. 232

  21. 21. 2327

  22. 22. 3437

Simplify each exponential expression in Exercises 2364.

  1. 23. x2 y

  2. 24. x y3

  3. 25. x0 y5

  4. 26. x7 y0

  5. 27. x3x7

  6. 28. x11x5

  7. 29. x5x10

  8. 30. x6x12

  9. 31. (x3)7

  10. 32. (x11)5

  11. 33. (x5)3

  12. 34. (x6)4

  13. 35. x14x7

  14. 36. x30x10

  15. 37. x14x7

  16. 38. x30x10

  17. 39. (8x3)2

  18. 40. (6x4)2

  19. 41. (4x)3

  20. 42. (6y)3

  21. 43. (3x2 y5)2

  22. 44. (3x4 y6)3

  23. 45. (3x4)(2x7)

  24. 46. (11x5)(9x12)

  25. 47. (9x3 y)(2x6 y4)

  26. 48. (5x4 y)(6x7 y11)

  27. 49. 8x20 2x4

  28. 50. 20x24 10x6

  29. 51. 25a13 b45a2 b3

  30. 52. 35a14 b67a7 b3

  31. 53. 14b7 7b14

  32. 54. 20b10 10b20

  33. 55. (4x3)2

  34. 56. (10x2) 3

  35. 57. 24x3 y532x7 y9

  36. 58. 10x4 y930x12 y3

  37. 59. (5x3y)2

  38. 60. (3x4y)3

  39. 61. (15a4 b25a10 b3)3

  40. 62. (30a14 b810a17 b2)3

  41. 63. (3a5 b212a3 b4)0

  42. 64. (4a5 b312a3 b5)0

In Exercises 6576, write each number in decimal notation without the use of exponents.

  1. 65. 3.8×102

  2. 66. 9.2×102

  3. 67. 6×104

  4. 68. 7×105

  5. 69. 7.16×106

  6. 70. 8.17×106

  7. 71. 7.9×101

  8. 72. 6.8×101

  9. 73. 4.15×103

  10. 74. 3.14×103

  11. 75. 6.00001×1010

  12. 76. 7.00001×1010

In Exercises 7786, write each number in scientific notation.

  1. 77. 32,000

  2. 78. 64,000

  3. 79. 638,000,000,000,000,000

  4. 80. 579,000,000,000,000,000

  5. 81. 5716

  6. 82. 3829

  7. 83. 0.0027

  8. 84. 0.0083

  9. 85. 0.00000000504

  10. 86. 0.00000000405

In Exercises 87106, perform the indicated computations. Write the answers in scientific notation. If necessary, round the decimal factor in your scientific notation answer to two decimal places.

  1. 87. (3×104)(2.1×103)

  2. 88. (2×104)(4.1×103)

  3. 89. (1.6×1015)(4×1011)

  4. 90. (1.4×1015)(3×1011)

  5. 91. (6.1×108)(2×104)

  6. 92. (5.1×108)(3×104)

  7. 93. (4.3×108)(6.2×104)

  8. 94. (8.2×108)(4.6×104)

  9. 95. 8.4×1084×105

  10. 96. 6.9×1083×105

  11. 97. 3.6×1049×102

  12. 98. 1.2×1042×102

  13. 99. 4.8×1022.4×106

  14. 100. 7.5×1022.5×106

  15. 101. 2.4×1024.8×106

  16. 102. 1.5×1023×106

  17. 103. 480,000,000,000 0.00012

  18. 104. 282,000,000,000 0.00141

  19. 105. 0.00072×0.003 0.00024

  20. 106. 66,000×0.001 0.003×0.002

Practice PLUS

In Exercises 107114, simplify each exponential expression. Assume that variables represent nonzero real numbers.

  1. 107. (x2 y)3(x2 y1)3

  2. 108. (xy2)2(x2y)3

  3. 109. (2x3yz6)(2x)5

  4. 110. (3x4 y z7)(3x)3

  5. 111. (x3y4z5x3y4z5)2

  6. 112. (x4 y5 z6x4 y5 z6)4

  7. 113. (21x2x1)2(2x4y3)2(16x3y3)0(2x3y5)2

  8. 114. (21x3y1)2(2x6y4)2(9x3y3)0(2x4y6)2

Application Exercises

The graph shows the cost, in billions of dollars, and the enrollment, in millions of people, for various federal social programs for a recent year. Use the numbers shown to solve Exercises 115117.

A bar graph titled, cost and enrolment for Federal social programs.

Source: Office of Management and Budget

  1. 115.

    1. Express the federal cost for Social Security in scientific notation.

    2. Express the enrollment in Social Security in scientific notation.

    3. Use your answers from parts (a) and (b) to determine the average yearly per person benefit for Social Security. Round to the nearest dollar and express the answer in scientific notation and in decimal notation.

    4. Use your decimal answer from part (c) to determine the average monthly per person benefit, expressed in decimal notation, for Social Security.

  2. 116.

    1. Express the federal cost for the food stamps program in scientific notation.

    2. Express the enrollment in the food stamps program in scientific notation.

    3. Use your answers from parts (a) and (b) to determine the average yearly per person benefit for the food stamps program. Round to the nearest dollar and express the answer in scientific notation and in decimal notation.

    4. Use your decimal answer from part (c) to determine the average monthly per person benefit, expressed in decimal notation, for the food stamps program.

  3. 117. Medicaid provides health insurance for the poor. Medicare provides health insurance for people 65 and older, as well as younger people who are disabled. Which program provides the greater yearly per person benefit? By how much, rounded to the nearest dollar?

We have seen that in May 2020 the U.S. national debt was $25.5 trillion. In Exercises 118120, you will use scientific notation to put a number like 25.5 trillion in perspective.

  1. 118.

    1. Express 25.5 trillion in scientific notation.

    2. Each year, Americans spend $254 billion on summer vacations. Express this number in scientific notation.

    3. Use your answers from parts (a) and (b) to determine how many years Americans can have free summer vacations for $25.5 trillion.

  1. 119.

    1. Express 25.5 trillion in scientific notation.

    2. Assume that four years of tuition, fees, and room and board at a public U.S. college cost approximately $60,000. Express this number in scientific notation.

    3. Use your answers from parts (a) and (b) to determine how many Americans could receive a free college education for $25.5 trillion.

  2. 120. In 2019, the United States government spent more than it had collected in taxes, resulting in a budget deficit of $984 billion.

    1. Express 984 billion in scientific notation.

    2. There are approximately 32,000,000 seconds in a year. Write this number in scienfic notation.

    3. Use your answers from parts (a) and (b) to determine approximately how many years is 984 billion seconds. (Note: 984 billion seconds would take us back in time to a period when Neanderthals were using stones to make tools.)

Explaining the Concepts

  1. 121. Describe what it means to raise a number to a power. In your description, include a discussion of the difference between 52 and (5)2.

  2. 122. Explain the product rule for exponents. Use 2325 in your explanation.

  3. 123. Explain the power rule for exponents. Use (32)4 in your explanation.

  4. 124. Explain the quotient rule for exponents. Use 58 52 in your explanation.

  5. 125. Why is (3x2)(2x5) not simplified? What must be done to simplify the expression?

  6. 126. How do you know if a number is written in scientific notation?

  7. 127. Explain how to convert from scientific to decimal notation and give an example.

  8. 128. Explain how to convert from decimal to scientific notation and give an example.

Critical Thinking Exercises

Make Sense? In Exercises 129132, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 129. There are many exponential expressions that are equal to 36x12, such as (6x6)2, (6x3)(6x9), 36(x3)9, and 62(x2)6.

  2. 130. If 52 is raised to the third power, the result is a number between 0 and 1.

  3. 131. The population of Colorado is approximately 4.6×1012.

  4. 132. I just finished reading a book that contained approximately 1.04×105 words.

In Exercises 133140, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 133. 42<43

  2. 134. 52>25

  3. 135. (2)4=24

  4. 136. 5252>2525

  5. 137. 534.7=5.347×103

  6. 138. 8×10304×105=2×1025

  7. 139. (7×105)+(2×103)=9×102

  8. 140. (4×103)+(3×102)=4.3×103

  9. 141. The mad Dr. Frankenstein has gathered enough bits and pieces (so to speak) for 21+22 of his creature-to-be. Write a fraction that represents the amount of his creature that must still be obtained.

  10. 142. If bA=MN, bC=M, and bD=N, what is the relationship among A, C, and D?

  11. 143. Our hearts beat approximately 70 times per minute. Express in scientific notation how many times the heart beats over a lifetime of 80 years. Round the decimal factor in your scientific notation answer to two decimal places.

Group Exercise

  1. 144. Putting Numbers into Perspective. A large number can be put into perspective by comparing it with another number. For example, we put the $25.5 trillion national debt in perspective (Example 6) by comparing this number to the number of U.S. citizens.

    For this project, each group member should consult an almanac, a newspaper, or the Internet to find a number greater than one million. Explain to other members of the group the context in which the large number is used. Express the number in scientific notation. Then put the number into perspective by comparing it with another number.

Preview Exercises

Exercises 145147 will help you prepare for the material covered in the next section.

  1. 145.

    1. Find 164.

    2. Find 164.

    3. Based on your answers to parts (a) and (b), what can you conclude?

  2. 146.

    1. Use a calculator to approximate 300 to two decimal places.

    2. Use a calculator to approximate 103 to two decimal places.

    3. Based on your answers to parts (a) and (b), what can you conclude?

  3. 147.

    1. Simplify: 21x+10x.

    2. Simplify: 212+102.

P.2: Exercise Set

P.2 Exercise Set

Practice Exercises

Evaluate each exponential expression in Exercises 122.

  1. 1. 522

  2. 2. 622

  3. 3. (2)6

  4. 4. (2)4

  5. 5. 26

  6. 6. 24

  7. 7. (3)0

  8. 8. (9)0

  9. 9. 30

  10. 10. 90

  11. 11. 43

  12. 12. 26

  13. 13. 2223

  14. 14. 3332

  15. 15. (22)3

  16. 16. (33)2

  17. 17. 2824

  18. 18. 3834

  19. 19. 333

  20. 20. 232

  21. 21. 2327

  22. 22. 3437

Simplify each exponential expression in Exercises 2364.

  1. 23. x2 y

  2. 24. x y3

  3. 25. x0 y5

  4. 26. x7 y0

  5. 27. x3x7

  6. 28. x11x5

  7. 29. x5x10

  8. 30. x6x12

  9. 31. (x3)7

  10. 32. (x11)5

  11. 33. (x5)3

  12. 34. (x6)4

  13. 35. x14x7

  14. 36. x30x10

  15. 37. x14x7

  16. 38. x30x10

  17. 39. (8x3)2

  18. 40. (6x4)2

  19. 41. (4x)3

  20. 42. (6y)3

  21. 43. (3x2 y5)2

  22. 44. (3x4 y6)3

  23. 45. (3x4)(2x7)

  24. 46. (11x5)(9x12)

  25. 47. (9x3 y)(2x6 y4)

  26. 48. (5x4 y)(6x7 y11)

  27. 49. 8x20 2x4

  28. 50. 20x24 10x6

  29. 51. 25a13 b45a2 b3

  30. 52. 35a14 b67a7 b3

  31. 53. 14b7 7b14

  32. 54. 20b10 10b20

  33. 55. (4x3)2

  34. 56. (10x2) 3

  35. 57. 24x3 y532x7 y9

  36. 58. 10x4 y930x12 y3

  37. 59. (5x3y)2

  38. 60. (3x4y)3

  39. 61. (15a4 b25a10 b3)3

  40. 62. (30a14 b810a17 b2)3

  41. 63. (3a5 b212a3 b4)0

  42. 64. (4a5 b312a3 b5)0

In Exercises 6576, write each number in decimal notation without the use of exponents.

  1. 65. 3.8×102

  2. 66. 9.2×102

  3. 67. 6×104

  4. 68. 7×105

  5. 69. 7.16×106

  6. 70. 8.17×106

  7. 71. 7.9×101

  8. 72. 6.8×101

  9. 73. 4.15×103

  10. 74. 3.14×103

  11. 75. 6.00001×1010

  12. 76. 7.00001×1010

In Exercises 7786, write each number in scientific notation.

  1. 77. 32,000

  2. 78. 64,000

  3. 79. 638,000,000,000,000,000

  4. 80. 579,000,000,000,000,000

  5. 81. 5716

  6. 82. 3829

  7. 83. 0.0027

  8. 84. 0.0083

  9. 85. 0.00000000504

  10. 86. 0.00000000405

In Exercises 87106, perform the indicated computations. Write the answers in scientific notation. If necessary, round the decimal factor in your scientific notation answer to two decimal places.

  1. 87. (3×104)(2.1×103)

  2. 88. (2×104)(4.1×103)

  3. 89. (1.6×1015)(4×1011)

  4. 90. (1.4×1015)(3×1011)

  5. 91. (6.1×108)(2×104)

  6. 92. (5.1×108)(3×104)

  7. 93. (4.3×108)(6.2×104)

  8. 94. (8.2×108)(4.6×104)

  9. 95. 8.4×1084×105

  10. 96. 6.9×1083×105

  11. 97. 3.6×1049×102

  12. 98. 1.2×1042×102

  13. 99. 4.8×1022.4×106

  14. 100. 7.5×1022.5×106

  15. 101. 2.4×1024.8×106

  16. 102. 1.5×1023×106

  17. 103. 480,000,000,000 0.00012

  18. 104. 282,000,000,000 0.00141

  19. 105. 0.00072×0.003 0.00024

  20. 106. 66,000×0.001 0.003×0.002

Practice PLUS

In Exercises 107114, simplify each exponential expression. Assume that variables represent nonzero real numbers.

  1. 107. (x2 y)3(x2 y1)3

  2. 108. (xy2)2(x2y)3

  3. 109. (2x3yz6)(2x)5

  4. 110. (3x4 y z7)(3x)3

  5. 111. (x3y4z5x3y4z5)2

  6. 112. (x4 y5 z6x4 y5 z6)4

  7. 113. (21x2x1)2(2x4y3)2(16x3y3)0(2x3y5)2

  8. 114. (21x3y1)2(2x6y4)2(9x3y3)0(2x4y6)2

Application Exercises

The graph shows the cost, in billions of dollars, and the enrollment, in millions of people, for various federal social programs for a recent year. Use the numbers shown to solve Exercises 115117.

A bar graph titled, cost and enrolment for Federal social programs.

Source: Office of Management and Budget

  1. 115.

    1. Express the federal cost for Social Security in scientific notation.

    2. Express the enrollment in Social Security in scientific notation.

    3. Use your answers from parts (a) and (b) to determine the average yearly per person benefit for Social Security. Round to the nearest dollar and express the answer in scientific notation and in decimal notation.

    4. Use your decimal answer from part (c) to determine the average monthly per person benefit, expressed in decimal notation, for Social Security.

  2. 116.

    1. Express the federal cost for the food stamps program in scientific notation.

    2. Express the enrollment in the food stamps program in scientific notation.

    3. Use your answers from parts (a) and (b) to determine the average yearly per person benefit for the food stamps program. Round to the nearest dollar and express the answer in scientific notation and in decimal notation.

    4. Use your decimal answer from part (c) to determine the average monthly per person benefit, expressed in decimal notation, for the food stamps program.

  3. 117. Medicaid provides health insurance for the poor. Medicare provides health insurance for people 65 and older, as well as younger people who are disabled. Which program provides the greater yearly per person benefit? By how much, rounded to the nearest dollar?

We have seen that in May 2020 the U.S. national debt was $25.5 trillion. In Exercises 118120, you will use scientific notation to put a number like 25.5 trillion in perspective.

  1. 118.

    1. Express 25.5 trillion in scientific notation.

    2. Each year, Americans spend $254 billion on summer vacations. Express this number in scientific notation.

    3. Use your answers from parts (a) and (b) to determine how many years Americans can have free summer vacations for $25.5 trillion.

  1. 119.

    1. Express 25.5 trillion in scientific notation.

    2. Assume that four years of tuition, fees, and room and board at a public U.S. college cost approximately $60,000. Express this number in scientific notation.

    3. Use your answers from parts (a) and (b) to determine how many Americans could receive a free college education for $25.5 trillion.

  2. 120. In 2019, the United States government spent more than it had collected in taxes, resulting in a budget deficit of $984 billion.

    1. Express 984 billion in scientific notation.

    2. There are approximately 32,000,000 seconds in a year. Write this number in scienfic notation.

    3. Use your answers from parts (a) and (b) to determine approximately how many years is 984 billion seconds. (Note: 984 billion seconds would take us back in time to a period when Neanderthals were using stones to make tools.)

Explaining the Concepts

  1. 121. Describe what it means to raise a number to a power. In your description, include a discussion of the difference between 52 and (5)2.

  2. 122. Explain the product rule for exponents. Use 2325 in your explanation.

  3. 123. Explain the power rule for exponents. Use (32)4 in your explanation.

  4. 124. Explain the quotient rule for exponents. Use 58 52 in your explanation.

  5. 125. Why is (3x2)(2x5) not simplified? What must be done to simplify the expression?

  6. 126. How do you know if a number is written in scientific notation?

  7. 127. Explain how to convert from scientific to decimal notation and give an example.

  8. 128. Explain how to convert from decimal to scientific notation and give an example.

Critical Thinking Exercises

Make Sense? In Exercises 129132, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 129. There are many exponential expressions that are equal to 36x12, such as (6x6)2, (6x3)(6x9), 36(x3)9, and 62(x2)6.

  2. 130. If 52 is raised to the third power, the result is a number between 0 and 1.

  3. 131. The population of Colorado is approximately 4.6×1012.

  4. 132. I just finished reading a book that contained approximately 1.04×105 words.

In Exercises 133140, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 133. 42<43

  2. 134. 52>25

  3. 135. (2)4=24

  4. 136. 5252>2525

  5. 137. 534.7=5.347×103

  6. 138. 8×10304×105=2×1025

  7. 139. (7×105)+(2×103)=9×102

  8. 140. (4×103)+(3×102)=4.3×103

  9. 141. The mad Dr. Frankenstein has gathered enough bits and pieces (so to speak) for 21+22 of his creature-to-be. Write a fraction that represents the amount of his creature that must still be obtained.

  10. 142. If bA=MN, bC=M, and bD=N, what is the relationship among A, C, and D?

  11. 143. Our hearts beat approximately 70 times per minute. Express in scientific notation how many times the heart beats over a lifetime of 80 years. Round the decimal factor in your scientific notation answer to two decimal places.

Group Exercise

  1. 144. Putting Numbers into Perspective. A large number can be put into perspective by comparing it with another number. For example, we put the $25.5 trillion national debt in perspective (Example 6) by comparing this number to the number of U.S. citizens.

    For this project, each group member should consult an almanac, a newspaper, or the Internet to find a number greater than one million. Explain to other members of the group the context in which the large number is used. Express the number in scientific notation. Then put the number into perspective by comparing it with another number.

Preview Exercises

Exercises 145147 will help you prepare for the material covered in the next section.

  1. 145.

    1. Find 164.

    2. Find 164.

    3. Based on your answers to parts (a) and (b), what can you conclude?

  2. 146.

    1. Use a calculator to approximate 300 to two decimal places.

    2. Use a calculator to approximate 103 to two decimal places.

    3. Based on your answers to parts (a) and (b), what can you conclude?

  3. 147.

    1. Simplify: 21x+10x.

    2. Simplify: 212+102.

Objective 1: Evaluate square roots

Square Roots

  1. Objective 1Evaluate square roots.

Watch Video

From our earlier work with exponents, we are aware that the square of both 5 and 5 is 25:

52=25and(5)2=25.

The reverse operation of squaring a number is finding the square root of a number. For example,

In general, if b2 =a, then b is a square root of a.

The symbol   is used to denote the nonnegative or principal square root of a number. For example,

The symbol   that we use to denote the principal square root is called a radical sign. The number under the radical sign is called the radicand. Together we refer to the radical sign and its radicand as a radical expression.

Radical a is labeled Radical expression. The square root symbol is labeled radical sign. a is labeled radicand.

Definition of the Principal Square Root

If a is a nonnegative real number, the nonnegative number b such that b2=a, denoted by b=a, is the principal square root of a.

The symbol    is used to denote the negative square root of a number. For example,

Example 1 Evaluating Square Roots

Evaluate:

  1. 64

  2. 49 

  3. 14

  4. 9+16

  5. 9+16.

Solution

  1. 64=8The principal square root of 64 is 8. Check: 82=64.

  2. 49=7The negative square root of 49 is 7. Check: (7)2=49.

  3. 14=12The principal square root of 14 is 12. Check: (12)2=14.

  4. 9+16=25First simplify the expression under the radical sign.=5Then take the principal square root of 25, which is 5.

  5. 9+16=3+49=3 because 32=9. 16=4 because 42=16.=7

Check Point 1

  • Evaluate:

    1. 81

    2. 9

    3. 125

    4. 36+64

    5. 36+64.

A number that is the square of a rational number is called a perfect square. All the radicands in Example 1 and Check Point 1 are perfect squares.

Let’s see what happens to the radical expression x if x is a negative number. Is the square root of a negative number a real number? For example, consider 25. Is there a real number whose square is 25? No. Thus, 25 is not a real number. In general, a square root of a negative number is not a real number.

If a number a is nonnegative (a0), then (a)2=a. For example,

(2)2=2, (3)2=3, (4)2=4, and(5 )2=5.
Objective 1: Evaluate square roots

Square Roots

  1. Objective 1Evaluate square roots.

Watch Video

From our earlier work with exponents, we are aware that the square of both 5 and 5 is 25:

52=25and(5)2=25.

The reverse operation of squaring a number is finding the square root of a number. For example,

In general, if b2 =a, then b is a square root of a.

The symbol   is used to denote the nonnegative or principal square root of a number. For example,

The symbol   that we use to denote the principal square root is called a radical sign. The number under the radical sign is called the radicand. Together we refer to the radical sign and its radicand as a radical expression.

Radical a is labeled Radical expression. The square root symbol is labeled radical sign. a is labeled radicand.

Definition of the Principal Square Root

If a is a nonnegative real number, the nonnegative number b such that b2=a, denoted by b=a, is the principal square root of a.

The symbol    is used to denote the negative square root of a number. For example,

Example 1 Evaluating Square Roots

Evaluate:

  1. 64

  2. 49 

  3. 14

  4. 9+16

  5. 9+16.

Solution

  1. 64=8The principal square root of 64 is 8. Check: 82=64.

  2. 49=7The negative square root of 49 is 7. Check: (7)2=49.

  3. 14=12The principal square root of 14 is 12. Check: (12)2=14.

  4. 9+16=25First simplify the expression under the radical sign.=5Then take the principal square root of 25, which is 5.

  5. 9+16=3+49=3 because 32=9. 16=4 because 42=16.=7

Check Point 1

  • Evaluate:

    1. 81

    2. 9

    3. 125

    4. 36+64

    5. 36+64.

A number that is the square of a rational number is called a perfect square. All the radicands in Example 1 and Check Point 1 are perfect squares.

Let’s see what happens to the radical expression x if x is a negative number. Is the square root of a negative number a real number? For example, consider 25. Is there a real number whose square is 25? No. Thus, 25 is not a real number. In general, a square root of a negative number is not a real number.

If a number a is nonnegative (a0), then (a)2=a. For example,

(2)2=2, (3)2=3, (4)2=4, and(5 )2=5.
Objective 3: Use the product rule to simplify square roots

The Product Rule for Square Roots

  1. Objective 3Use the product rule to simplify square roots.

Watch Video

A rule for multiplying square roots can be generalized by comparing 254 and 254. Notice that

25 4=52=10and254=100 =10.

Because we obtain 10 in both situations, the original radical expressions must be equal. That is,

254=254.

This result is a special case of the product rule for square roots that can be generalized as follows:

The Product Rule for Square Roots

If a and b represent nonnegative real numbers, then

The image shows the product rule for square roots.

A square root is simplified when its radicand has no factors other than 1 that are perfect squares. For example, 500 is not simplified because it can be expressed as 1005 and 100 is a perfect square. Example 2 shows how the product rule is used to remove from the square root any perfect squares that occur as factors.

Example 2 Using the Product Rule to Simplify Square Roots

Simplify:

  1. 500

  2. 6x3x.

Solution

  1.  

    500=1005Factor 500. 100 is the greatest perfect square factor.=1005Use the product rule:  ab=ab.=105Write  100   as 10. We read  105 as ten times the squareroot of 5.
  2. We can simplify 6x3x using the product rule only if 6x and 3x represent nonnegative real numbers. Thus, x0.

    6x3x=6x3xUse the product rule:  ab=ab.=18x2Multiply in the radicand.=9x22Factor 18. 9 is the greatest perfect square factor.=9x2 2Use the product rule:  ab=ab.=9 x2 2Use the product rule to write  9x 2  as the productof two square roots.=3x2x 2=|x |=x  because  x0.

Check Point 2

  • Simplify:

    1. 75

    2. 5x10x.

Objective 4: Use the quotient rule to simplify square roots

The Quotient Rule for Square Roots

  1. Objective 4Use the quotient rule to simplify square roots.

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Another property for square roots involves division.

The Quotient Rule for Square Roots

If a and b represent nonnegative real numbers and b0, then

The quotient rule for square roots.

Example 3 Using the Quotient Rule to Simplify Square Roots

Simplify:

  1. 1009

  2. 48x36x.

Solution

  1. 1009=1009=103

  2. We can simplify the quotient of 48x3 and 6x using the quotient rule only if 48x3 and 6x represent nonnegative real numbers and 6x0. Thus, x>0.

    Simplifying start fraction the square root of start expression 48 x cubed end expression over the square root of start expression 6 x end expression end fraction.

Check Point 3

  • Simplify:

    1. 2516

    2. 150x32x.

Objective 5: Add and subtract square roots

Adding and Subtracting Square Roots

  1. Objective 5Add and subtract square roots.

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Two or more square roots can be combined using the distributive property provided that they have the same radicand. Such radicals are called like radicals. For example,

7 radical 11 plus 6 radical 11 = left parenthesis 7 + 6 right parenthesis radical 11 = 13 radical 11. A note pointing at the square roots sign reads 7 square roots of 11 plus 6 square roots of 11 result in 13 square roots of 11.

Example 4 Adding and Subtracting Like Radicals

Add or subtract as indicated:

  1. 72+52

  2. 5x75x.

Solution

  1.  

     72+52=(7+5)2Apply the distributive property.=122Simplify.
  2.  

     5x75x=15x75xWrite 5x as 15x.=(17)5xApply the distributive property.=65xSimplify.

Check Point 4

  • Add or subtract as indicated:

    1. 813+913

    2. 17x2017x.

In some cases, radicals can be combined once they have been simplified. For example, to add 2 and 8, we can write 8 as 42 because 4 is a perfect square factor of 8.

2+8=2+42=12+22=(1+2)2=32

Example 5 Combining Radicals That First Require Simplification

Add or subtract as indicated:

  1. 73+12

  2. 450x632x.

Solution

  1.  

    73+12=73+43Split 12 into two factors such that one is aperfect square.=73+2343=43=23=(7+2)3Apply the distributive property.You will find that this step is usually done mentally.=93Simplify.
  2.  

    450x632x=4252x6162x25 is the greatest perfect square factor of  50x  and 16 isthe greatest perfect square factor of  32x.=452x642x252x=252x=52x  and162x=162x=42x.000=202x242xMultiply:  45=20  and  64=24.=(2024)2xApply the distributive property.=42xSimplify.

Check Point 5

  • Add or subtract as indicated:

    1. 527+12

    2. 618x48x.

Objective 6: Rationalize denominators

Rationalizing Denominators

  1. Objective 6Rationalize denominators.

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The calculator screen in Figure P.11 shows approximate values for 13 and 33 . The two approximations are the same. This is not a coincidence:

Start fraction 1 over radical 3 end fraction = start fraction 1 over radical 3 end fraction times start fraction radical 3 over radical 3 end fraction = start fraction radical 3 over radical 9 end fraction = radical 3 thirds.
P.3-59 Full Alternative Text

Figure P.11 The calculator screen shows approximate values for 13 and 33 .

A graphing calculator screen displays two inputs and the corresponding outputs. Input 1. Start fraction 1 over radical 3 end fraction. Output 1. 0.5773502692. Input 2. Start fraction radical 3 over 3 end fraction. Output 2. 0.5773502692.

This process involves rewriting a radical expression as an equivalent expression in which the denominator no longer contains any radicals. The process is called rationalizing the denominator. If the denominator consists of the square root of a natural number that is not a perfect square, multiply the numerator and the denominator by the smallest number that produces the square root of a perfect square in the denominator.

Example 6 Rationalizing Denominators

Rationalize the denominator:

  1. 156

  2. 128 .

Solution

  1. If we multiply the numerator and the denominator of 156 by 6, the denominator becomes 66=36=6. Therefore, we multiply by 1, choosing 66 for 1.

    The process to simplify start fraction 15 over radical 6 end fraction.
  2. The smallest number that will produce the square root of a perfect square in the denominator of 12 8  is 2 , because 82=16=4. We multiply by 1, choosing 22  for 1.

    128=12822=12216=1224=32 

Check Point 6

  • Rationalize the denominator:

    1. 53

    2. 612.

Radical expressions that involve the sum and difference of the same two terms are called conjugates. Thus,

a+bandab

are conjugates. Conjugates are used to rationalize denominators because the product of such a pair contains no radicals:

The process to simplify left parenthesis radical a + radical b right parenthesis left parenthesis radical a minus radical b right parenthesis.
P.3-61 Full Alternative Text

Multiplying Conjugates

(a+b)(ab)=(a)2(b)2=ab

How can we rationalize a denominator if the denominator contains two terms with one or more square roots? Multiply the numerator and the denominator by the conjugate of the denominator. Here are three examples of such expressions:

Three expressions with the conjugate of the denominators are mentioned.
P.3-62 Full Alternative Text

The product of the denominator and its conjugate is found using the formula

(a+b)(ab)=(a)2(b)2=ab.

The simplified product will not contain a radical.

Example 7 Rationalizing a Denominator Containing Two Terms

Rationalize the denominator: 75+3.

Solution

The conjugate of the denominator is 53. If we multiply the numerator and denominator by 53, the simplified denominator will not contain a radical. Therefore, we multiply by 1, choosing 5353 for 1.

The process to simplify start fraction 7 over 5 + radical 3 end fraction.

Check Point 7

  • Rationalize the denominator: 84+5.

Objective 6: Rationalize denominators

Rationalizing Denominators

  1. Objective 6Rationalize denominators.

Watch Video

The calculator screen in Figure P.11 shows approximate values for 13 and 33 . The two approximations are the same. This is not a coincidence:

Start fraction 1 over radical 3 end fraction = start fraction 1 over radical 3 end fraction times start fraction radical 3 over radical 3 end fraction = start fraction radical 3 over radical 9 end fraction = radical 3 thirds.
P.3-59 Full Alternative Text

Figure P.11 The calculator screen shows approximate values for 13 and 33 .

A graphing calculator screen displays two inputs and the corresponding outputs. Input 1. Start fraction 1 over radical 3 end fraction. Output 1. 0.5773502692. Input 2. Start fraction radical 3 over 3 end fraction. Output 2. 0.5773502692.

This process involves rewriting a radical expression as an equivalent expression in which the denominator no longer contains any radicals. The process is called rationalizing the denominator. If the denominator consists of the square root of a natural number that is not a perfect square, multiply the numerator and the denominator by the smallest number that produces the square root of a perfect square in the denominator.

Example 6 Rationalizing Denominators

Rationalize the denominator:

  1. 156

  2. 128 .

Solution

  1. If we multiply the numerator and the denominator of 156 by 6, the denominator becomes 66=36=6. Therefore, we multiply by 1, choosing 66 for 1.

    The process to simplify start fraction 15 over radical 6 end fraction.
  2. The smallest number that will produce the square root of a perfect square in the denominator of 12 8  is 2 , because 82=16=4. We multiply by 1, choosing 22  for 1.

    128=12822=12216=1224=32 

Check Point 6

  • Rationalize the denominator:

    1. 53

    2. 612.

Radical expressions that involve the sum and difference of the same two terms are called conjugates. Thus,

a+bandab

are conjugates. Conjugates are used to rationalize denominators because the product of such a pair contains no radicals:

The process to simplify left parenthesis radical a + radical b right parenthesis left parenthesis radical a minus radical b right parenthesis.
P.3-61 Full Alternative Text

Multiplying Conjugates

(a+b)(ab)=(a)2(b)2=ab

How can we rationalize a denominator if the denominator contains two terms with one or more square roots? Multiply the numerator and the denominator by the conjugate of the denominator. Here are three examples of such expressions:

Three expressions with the conjugate of the denominators are mentioned.
P.3-62 Full Alternative Text

The product of the denominator and its conjugate is found using the formula

(a+b)(ab)=(a)2(b)2=ab.

The simplified product will not contain a radical.

Example 7 Rationalizing a Denominator Containing Two Terms

Rationalize the denominator: 75+3.

Solution

The conjugate of the denominator is 53. If we multiply the numerator and denominator by 53, the simplified denominator will not contain a radical. Therefore, we multiply by 1, choosing 5353 for 1.

The process to simplify start fraction 7 over 5 + radical 3 end fraction.

Check Point 7

  • Rationalize the denominator: 84+5.

Objective 7: Evaluate and perform operations with higher roots

Other Kinds of Roots

  1. Objective 7Evaluate and perform operations with higher roots.

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We define the principal nth root of a real number a, symbolized by an, as follows:

Definition of the Principal nth Root of a Real Number

an=b means that bn=a.

If n, the index, is even, then a is nonnegative (a0) and b is also nonnegative (b0). If n is odd, a and b can be any real numbers.

For example,

643=4because43=64and325=2because (2)5=32.

The same vocabulary that we learned for square roots applies to nth roots. The symbol  n is called a radical and the expression under the radical is called the radicand.

A number that is the nth power of a rational number is called a perfect nth power. For example, 8 is a perfect third power, or perfect cube, because 8=23. Thus, 83=233=2. In general, one of the following rules can be used to find the nth root of a perfect nth power:

Finding nth Roots of Perfect nth Powers

If n is odd, ann=a.
If n is even, ann=|a|.

For example,

Two examples that absolute value is not needed with odd roots, but is necessary with even roots.
P.3-64 Full Alternative Text

The Product and Quotient Rules for Other Roots

The product and quotient rules apply to cube roots, fourth roots, and all higher roots.

The Product and Quotient Rules for nth Roots

For all real numbers a and b, where the indicated roots represent real numbers,

Examples of product and quotient rules for nth roots.

Example 8 Simplifying, Multiplying, and Dividing Higher Roots

Simplify:

  1. 243

  2. 84 44

  3. 8116.4

Solution

  1.  

    243=833Find the greatest perfect cube that is a factor of 24. 23=8, so8 is a perfect cube and is the greatest perfect cube factor of 24.=8333abn=anbn=23383=2
  2.  

    8444=844anbn=abn =324Find the greatest perfect fourth power that is a factor of 32.=1624  24=16,  so 16 is a perfect fourth power and is the greatestperfect fourth power that is a factor of 32.=16424abn=anbn =224164=2

  3.  

     81164=814164abn=anbn=32 814=3  because  34=81  and  164=2  because  24=16.

Check Point 8

  • Simplify:

    1. 403

    2. 8585

    3. 125273.

We have seen that adding and subtracting square roots often involves simplifying terms. The same idea applies to adding and subtracting higher roots.

Example 9 Combining Cube Roots

Subtract: 51631123.

Solution

51631123=58231123Factor 16. 8 is the greatest perfect cube factor: 23=8  and  83=2.=52231123823=8323=223=10231123Multiply:  52=10.=(1011)23Apply the distributive property.=123 or23Simplify.

Check Point 9

  • Subtract: 3813433.

Objective 7: Evaluate and perform operations with higher roots

Other Kinds of Roots

  1. Objective 7Evaluate and perform operations with higher roots.

Watch Video

We define the principal nth root of a real number a, symbolized by an, as follows:

Definition of the Principal nth Root of a Real Number

an=b means that bn=a.

If n, the index, is even, then a is nonnegative (a0) and b is also nonnegative (b0). If n is odd, a and b can be any real numbers.

For example,

643=4because43=64and325=2because (2)5=32.

The same vocabulary that we learned for square roots applies to nth roots. The symbol  n is called a radical and the expression under the radical is called the radicand.

A number that is the nth power of a rational number is called a perfect nth power. For example, 8 is a perfect third power, or perfect cube, because 8=23. Thus, 83=233=2. In general, one of the following rules can be used to find the nth root of a perfect nth power:

Finding nth Roots of Perfect nth Powers

If n is odd, ann=a.
If n is even, ann=|a|.

For example,

Two examples that absolute value is not needed with odd roots, but is necessary with even roots.
P.3-64 Full Alternative Text

The Product and Quotient Rules for Other Roots

The product and quotient rules apply to cube roots, fourth roots, and all higher roots.

The Product and Quotient Rules for nth Roots

For all real numbers a and b, where the indicated roots represent real numbers,

Examples of product and quotient rules for nth roots.

Example 8 Simplifying, Multiplying, and Dividing Higher Roots

Simplify:

  1. 243

  2. 84 44

  3. 8116.4

Solution

  1.  

    243=833Find the greatest perfect cube that is a factor of 24. 23=8, so8 is a perfect cube and is the greatest perfect cube factor of 24.=8333abn=anbn=23383=2
  2.  

    8444=844anbn=abn =324Find the greatest perfect fourth power that is a factor of 32.=1624  24=16,  so 16 is a perfect fourth power and is the greatestperfect fourth power that is a factor of 32.=16424abn=anbn =224164=2

  3.  

     81164=814164abn=anbn=32 814=3  because  34=81  and  164=2  because  24=16.

Check Point 8

  • Simplify:

    1. 403

    2. 8585

    3. 125273.

We have seen that adding and subtracting square roots often involves simplifying terms. The same idea applies to adding and subtracting higher roots.

Example 9 Combining Cube Roots

Subtract: 51631123.

Solution

51631123=58231123Factor 16. 8 is the greatest perfect cube factor: 23=8  and  83=2.=52231123823=8323=223=10231123Multiply:  52=10.=(1011)23Apply the distributive property.=123 or23Simplify.

Check Point 9

  • Subtract: 3813433.

Objective 8: Understand and use rational exponents

Rational Exponents

  1. Objective 8Understand and use rational exponents.

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We define rational exponents so that their properties are the same as the properties for integer exponents. For example, we know that exponents are multiplied when an exponential expression is raised to a power. For this to be true,

(712)2=7122=71=7.

We also know that

(7)2=77=49=7.

Can you see that the square of both 712 and 7 is 7? It is reasonable to conclude that

712means7.

We can generalize the fact that 712 means 7  with the following definition:

The Definition of a1n

If an represents a real number, where n2 is an integer, then

a to the power of start fraction 1 over n end fraction = the nth root of a. A note pointing at the n of the expression reads the denominator of the rational exponent is the radical’s index.

Furthermore,

a1n=1a1n=1an, a0.

Example 10 Using the Definition of a1n

Simplify:

  1. 6412

  2. 12513

  3. 1614

  4. (27)13

  5. 6413.

Solution

  1. 6412=64=8

  2.  

    125 to the 1 third power = the cube root of 125 = 5. A note pointing at 3 reads, the denominator is the index.
  3.  

    Negative 16 to the 1 fourth power = negative left parenthesis the fourth root of 16 right parenthesis = negative 2. A note pointing at the negative sign of the expression reads, the base is 16 and the negative sign is not affected by the exponent.
  4.  

    Left parenthesis negative 27 right parenthesis to the 1 third power = the cube root of negative 27 = negative 3. A note pointing at the expression reads, Parentheses show that the base is negative 27 and that the negative sign is affected by the exponent.
  5.  

    6413=16413=1643=14

Check Point 10

  • Simplify:

    1. 2512

    2. 813

    3. 8114

    4. (8)13

    5. 2713.

In Example 10 and Check Point 10, each rational exponent had a numerator of 1. If the numerator is some other integer, we still want to multiply exponents when raising a power to a power. For this reason,

a to the 2 thirds power can be expanded in two ways.
P.3-71 Full Alternative Text

Thus,

a23=(a3)2=a23.

Do you see that the denominator, 3, of the rational exponent is the same as the index of the radical? The numerator, 2, of the rational exponent serves as an exponent in each of the two radical forms. We generalize these ideas with the following definition:

The Definition of amn

If an  represents a real number and mn is a positive rational number, n2, then

amn=(an)m.

Also,

amn=amn.

Furthermore, if amn is a nonzero real number, then

amn=1amn.

The first form of the definition of amn, shown again below, involves taking the root first. This form is often preferable because smaller numbers are involved. Notice that the rational exponent consists of two parts, indicated by the following voice balloons:

a to the power of start fraction m over n end fraction = left parenthesis the nth root of a right parenthesis to the power of m. A note pointing at m reads, the numerator, m, is the exponent. A note pointing at n reads, the denominator, n, is the radical's index.

Example 11 Using the Definition of amn

Simplify:

  1. 2723

  2. 932

  3. 8134.

Solution

  1. 2723=(273)2=32=9

  2. 932=(9)3=33=27

  3. 8134=18134=1(814)3=133=127

Check Point 11

  • Simplify:

    1. 2743

    2. 432

    3. 3225.

Properties of exponents can be applied to expressions containing rational exponents.

Example 12 Simplifying Expressions with Rational Exponents

Simplify using properties of exponents:

  1. (5x12)(7x34)

  2. 32x5316x34.

Solution

  1.  

    (5x12)(7x34)=57x12x34Group numerical factors and group variable factorswith the same base.=35x12+34When multiplying expressions with the same base,add the exponents.=35x5412+34=24+34=54
  2.  

    32x5316x34=(3216)(x53x34)Group numerical factors and group variable factorswith the same base.=2x5334When dividing expressions with the same base,subtract the exponents.=2x11125334=2012912=1112

Check Point 12

  • Simplify using properties of exponents:

    1. (2x43)(5x83)

    2. 20x45x32.

Rational exponents are sometimes useful for simplifying radicals by reducing the index.

Example 13 Reducing the Index of a Radical

Simplify: x39.

Solution

x39=x39=x13=x3

Check Point 13

  • Simplify: x36.

Objective 8: Understand and use rational exponents

Rational Exponents

  1. Objective 8Understand and use rational exponents.

Watch Video

We define rational exponents so that their properties are the same as the properties for integer exponents. For example, we know that exponents are multiplied when an exponential expression is raised to a power. For this to be true,

(712)2=7122=71=7.

We also know that

(7)2=77=49=7.

Can you see that the square of both 712 and 7 is 7? It is reasonable to conclude that

712means7.

We can generalize the fact that 712 means 7  with the following definition:

The Definition of a1n

If an represents a real number, where n2 is an integer, then

a to the power of start fraction 1 over n end fraction = the nth root of a. A note pointing at the n of the expression reads the denominator of the rational exponent is the radical’s index.

Furthermore,

a1n=1a1n=1an, a0.

Example 10 Using the Definition of a1n

Simplify:

  1. 6412

  2. 12513

  3. 1614

  4. (27)13

  5. 6413.

Solution

  1. 6412=64=8

  2.  

    125 to the 1 third power = the cube root of 125 = 5. A note pointing at 3 reads, the denominator is the index.
  3.  

    Negative 16 to the 1 fourth power = negative left parenthesis the fourth root of 16 right parenthesis = negative 2. A note pointing at the negative sign of the expression reads, the base is 16 and the negative sign is not affected by the exponent.
  4.  

    Left parenthesis negative 27 right parenthesis to the 1 third power = the cube root of negative 27 = negative 3. A note pointing at the expression reads, Parentheses show that the base is negative 27 and that the negative sign is affected by the exponent.
  5.  

    6413=16413=1643=14

Check Point 10

  • Simplify:

    1. 2512

    2. 813

    3. 8114

    4. (8)13

    5. 2713.

In Example 10 and Check Point 10, each rational exponent had a numerator of 1. If the numerator is some other integer, we still want to multiply exponents when raising a power to a power. For this reason,

a to the 2 thirds power can be expanded in two ways.
P.3-71 Full Alternative Text

Thus,

a23=(a3)2=a23.

Do you see that the denominator, 3, of the rational exponent is the same as the index of the radical? The numerator, 2, of the rational exponent serves as an exponent in each of the two radical forms. We generalize these ideas with the following definition:

The Definition of amn

If an  represents a real number and mn is a positive rational number, n2, then

amn=(an)m.

Also,

amn=amn.

Furthermore, if amn is a nonzero real number, then

amn=1amn.

The first form of the definition of amn, shown again below, involves taking the root first. This form is often preferable because smaller numbers are involved. Notice that the rational exponent consists of two parts, indicated by the following voice balloons:

a to the power of start fraction m over n end fraction = left parenthesis the nth root of a right parenthesis to the power of m. A note pointing at m reads, the numerator, m, is the exponent. A note pointing at n reads, the denominator, n, is the radical's index.

Example 11 Using the Definition of amn

Simplify:

  1. 2723

  2. 932

  3. 8134.

Solution

  1. 2723=(273)2=32=9

  2. 932=(9)3=33=27

  3. 8134=18134=1(814)3=133=127

Check Point 11

  • Simplify:

    1. 2743

    2. 432

    3. 3225.

Properties of exponents can be applied to expressions containing rational exponents.

Example 12 Simplifying Expressions with Rational Exponents

Simplify using properties of exponents:

  1. (5x12)(7x34)

  2. 32x5316x34.

Solution

  1.  

    (5x12)(7x34)=57x12x34Group numerical factors and group variable factorswith the same base.=35x12+34When multiplying expressions with the same base,add the exponents.=35x5412+34=24+34=54
  2.  

    32x5316x34=(3216)(x53x34)Group numerical factors and group variable factorswith the same base.=2x5334When dividing expressions with the same base,subtract the exponents.=2x11125334=2012912=1112

Check Point 12

  • Simplify using properties of exponents:

    1. (2x43)(5x83)

    2. 20x45x32.

Rational exponents are sometimes useful for simplifying radicals by reducing the index.

Example 13 Reducing the Index of a Radical

Simplify: x39.

Solution

x39=x39=x13=x3

Check Point 13

  • Simplify: x36.

Objective 8: Understand and use rational exponents

Rational Exponents

  1. Objective 8Understand and use rational exponents.

Watch Video

We define rational exponents so that their properties are the same as the properties for integer exponents. For example, we know that exponents are multiplied when an exponential expression is raised to a power. For this to be true,

(712)2=7122=71=7.

We also know that

(7)2=77=49=7.

Can you see that the square of both 712 and 7 is 7? It is reasonable to conclude that

712means7.

We can generalize the fact that 712 means 7  with the following definition:

The Definition of a1n

If an represents a real number, where n2 is an integer, then

a to the power of start fraction 1 over n end fraction = the nth root of a. A note pointing at the n of the expression reads the denominator of the rational exponent is the radical’s index.

Furthermore,

a1n=1a1n=1an, a0.

Example 10 Using the Definition of a1n

Simplify:

  1. 6412

  2. 12513

  3. 1614

  4. (27)13

  5. 6413.

Solution

  1. 6412=64=8

  2.  

    125 to the 1 third power = the cube root of 125 = 5. A note pointing at 3 reads, the denominator is the index.
  3.  

    Negative 16 to the 1 fourth power = negative left parenthesis the fourth root of 16 right parenthesis = negative 2. A note pointing at the negative sign of the expression reads, the base is 16 and the negative sign is not affected by the exponent.
  4.  

    Left parenthesis negative 27 right parenthesis to the 1 third power = the cube root of negative 27 = negative 3. A note pointing at the expression reads, Parentheses show that the base is negative 27 and that the negative sign is affected by the exponent.
  5.  

    6413=16413=1643=14

Check Point 10

  • Simplify:

    1. 2512

    2. 813

    3. 8114

    4. (8)13

    5. 2713.

In Example 10 and Check Point 10, each rational exponent had a numerator of 1. If the numerator is some other integer, we still want to multiply exponents when raising a power to a power. For this reason,

a to the 2 thirds power can be expanded in two ways.
P.3-71 Full Alternative Text

Thus,

a23=(a3)2=a23.

Do you see that the denominator, 3, of the rational exponent is the same as the index of the radical? The numerator, 2, of the rational exponent serves as an exponent in each of the two radical forms. We generalize these ideas with the following definition:

The Definition of amn

If an  represents a real number and mn is a positive rational number, n2, then

amn=(an)m.

Also,

amn=amn.

Furthermore, if amn is a nonzero real number, then

amn=1amn.

The first form of the definition of amn, shown again below, involves taking the root first. This form is often preferable because smaller numbers are involved. Notice that the rational exponent consists of two parts, indicated by the following voice balloons:

a to the power of start fraction m over n end fraction = left parenthesis the nth root of a right parenthesis to the power of m. A note pointing at m reads, the numerator, m, is the exponent. A note pointing at n reads, the denominator, n, is the radical's index.

Example 11 Using the Definition of amn

Simplify:

  1. 2723

  2. 932

  3. 8134.

Solution

  1. 2723=(273)2=32=9

  2. 932=(9)3=33=27

  3. 8134=18134=1(814)3=133=127

Check Point 11

  • Simplify:

    1. 2743

    2. 432

    3. 3225.

Properties of exponents can be applied to expressions containing rational exponents.

Example 12 Simplifying Expressions with Rational Exponents

Simplify using properties of exponents:

  1. (5x12)(7x34)

  2. 32x5316x34.

Solution

  1.  

    (5x12)(7x34)=57x12x34Group numerical factors and group variable factorswith the same base.=35x12+34When multiplying expressions with the same base,add the exponents.=35x5412+34=24+34=54
  2.  

    32x5316x34=(3216)(x53x34)Group numerical factors and group variable factorswith the same base.=2x5334When dividing expressions with the same base,subtract the exponents.=2x11125334=2012912=1112

Check Point 12

  • Simplify using properties of exponents:

    1. (2x43)(5x83)

    2. 20x45x32.

Rational exponents are sometimes useful for simplifying radicals by reducing the index.

Example 13 Reducing the Index of a Radical

Simplify: x39.

Solution

x39=x39=x13=x3

Check Point 13

  • Simplify: x36.

P.3: Concept and Vocabulary Check

P.3 Concept and Vocabulary Check

Fill in each blank so that the resulting statement is true.

  1. C1. The symbol   is used to denote the nonnegative, or _______, square root of a number.

  2. C2. 64=8 because _____ =64.

  3. C3. a2= _______

  4. C4. The product rule for square roots states that if a and b are nonnegative, then ab= _______.

  5. C5. The quotient rule for square roots states that if a and b are nonnegative and b0, then ab= _______.

  6. C6. 83+103= _______

  7. C7. 3+75=3+253=3+_____3= _______

  8. C8. The conjugate of 7+3 is ______.

  9. C9. We rationalize the denominator of 5102 by multiplying the numerator and denominator by ________.

  10. C10. In the expression 643, the number 3 is called the ______ and the number 64 is called the _______.

  11. C11. 325=2 because _______ =32.

  12. C12. If n is odd, ann= _______.

    If n is even, ann= _______.

  13. C13. a1n= _______

  14. C14. 1634=(164)3=(__)3= ____

P.3: Exercise Set

P.3 Exercise Set

Practice Exercises

Evaluate each expression in Exercises 112, or indicate that the root is not a real number.

  1. 1. 36

  2. 2. 25

  3. 3. 36

  4. 4. 25

  5. 5. 36

  6. 6. 25

  7. 7. 2516

  8. 8. 144+25

  9. 9. 2516

  10. 10. 144+25

  11. 11. (13)2

  12. 12. (17)2

Use the product rule to simplify the expressions in Exercises 1322. In Exercises 1722, assume that variables represent nonnegative real numbers.

  1. 13. 50

  2. 14. 27

  3. 15. 45x2

  4. 16. 125x2

  5. 17. 2x6x

  6. 18. 10x 8x

  7. 19. x3

  8. 20. y3

  9. 21. 2x26x

  10. 22. 6x3x2

Use the quotient rule to simplify the expressions in Exercises 2332. Assume that x>0.

  1. 23. 181

  2. 24. 149

  3. 25. 4916

  4. 26. 1219

  5. 27. 48x33x

  6. 28. 72x38x

  7. 29. 150x43x

  8. 30. 24x43x

  9. 31. 200x310x1

  10. 32. 500x310x1

In Exercises 3344, add or subtract terms whenever possible.

  1. 33. 73+63

  2. 34. 85+115

  3. 35. 617x817x

  4. 36. 413x613x

  5. 37. 8+32

  6. 38. 20+65

  7. 39. 50x8x

  8. 40. 63x28x

  9. 41. 318+550

  10. 42. 412275

  11. 43. 3832+37275

  12. 44. 35422496+463

In Exercises 4554, rationalize the denominator.

  1. 45. 17

  2. 46. 210

  3. 47. 25

  4. 48. 73

  5. 49. 133+11

  6. 50. 33+7

  7. 51. 752

  8. 52. 531

  9. 53. 65+3

  10. 54. 1173

Evaluate each expression in Exercises 5566, or indicate that the root is not a real number.

  1. 55. 1253

  2. 56. 83

  3. 57. 83

  4. 58. 1253

  5. 59. 164

  6. 60. 814

  7. 61. (3) 44

  8. 62. (2) 44

  9. 63. (3)55

  10. 64. (2)55

  11. 65. 1325

  12. 66. 1646

Simplify the radical expressions in Exercises 6774, if possible.

  1. 67. 323

  2. 68. 1503

  3. 69. x43

  4. 70. x53

  5. 71. 9363

  6. 72. 12343

  7. 73. 64x652x5

  8. 74. 162x542x4

In Exercises 7582, add or subtract terms whenever possible.

  1. 75. 425+325

  2. 76. 635+235

  3. 77. 5163+543

  4. 78. 3243+813

  5. 79. 54xy33y128x3

  6. 80. 24xy33y81x3

  7. 81. 2+83

  8. 82. 3+153

In Exercises 8390, evaluate each expression without using a calculator.

  1. 83. 3612

  2. 84. 12112

  3. 85. 813

  4. 86. 2713

  5. 87. 12523

  6. 88. 823

  7. 89. 3245

  8. 90. 1652

In Exercises 91100, simplify using properties of exponents.

  1. 91. (7x13) (2x14)

  2. 92. (3x23) (4x34)

  3. 93. 20x125x14

  4. 94. 72x349x13

  5. 95. (x23)3

  6. 96. (x45)5

  7. 97. (25x4y6)12

  8. 98. (125x9y6)13

  9. 99. (3y14)3y112

  10. 100. (2y15)4y310

In Exercises 101108, simplify by reducing the index of the radical.

  1. 101. 524

  2. 102. 724

  3. 103. x63

  4. 104. x124

  5. 105. x46

  6. 106. x69

  7. 107. x6 y39

  8. 108. x4y812

Practice PLUS

In Exercises 109110, evaluate each expression.

  1. 109. 164+6253

  2. 110. 169+9+10003+21633

In Exercises 111114, simplify each expression. Assume that all variables represent positive numbers.

  1. 111. (49x2y4)12(xy12)

  2. 112. (8x6y3)13(x56y13)6

  3. 113. (x54y13x34)6

  4. 114. (x12y74y54)4

Application Exercises

  1. 115. Do you expect to pay more taxes than were withheld? Would you be surprised to know that the percentage of taxpayers who receive a refund and the percentage of taxpayers who pay more taxes vary according to age? The formula

    P=x(13+x)5x

    models the percentage, P, of taxpayers who are x years old who must pay more taxes.

    1. What percentage of 25-year-olds must pay more taxes?

    2. Rewrite the formula by rationalizing the denominator.

    3. Use the rationalized form of the formula from part (b) to find the percentage of 25-year-olds who must pay more taxes. Do you get the same answer as you did in part (a)? If so, does this prove that you correctly rationalized the denominator? Explain.

  2. 116. America is getting older. The graph shows the elderly U.S. population for ages 65–84 and for ages 85 and older in 2010, with projections for 2020 and beyond.

    A bar graph titled projected elderly United States population.

    Source: U.S. Census Bureau

    The formula E=5x+34.1 models the projected number of Americans ages 65–84, E, in millions, x years after 2010.

    1. Use the formula to find the projected increase in the number of Americans ages 65–84, in millions, from 2020 to 2050. Express this difference in simplified radical form.

    2. Use a calculator and write your answer in part (a) to the nearest tenth. Does this rounded decimal overestimate or underestimate the difference in the projected data shown by the bar graph? By how much?

  3. 117. The early Greeks believed that the most pleasing of all rectangles were golden rectangles, whose ratio of width to height is

    wh=251.

    The Parthenon at Athens fits into a golden rectangle once the triangular pediment is reconstructed.

    Rationalize the denominator of the given ratio, called the golden ratio. Then use a calculator and find the ratio of width to height, correct to the nearest hundredth, in golden rectangles.

  4. 118. Use Einstein’s special-relativity equation

    Ra=Rf1(vc)2,

    described in the Blitzer Bonus here, to solve this exercise. You are moving at 90% of the speed of light. Substitute 0.9c for v, your velocity, in the equation. What is your aging rate, correct to two decimal places, relative to a friend on Earth? If you are gone for 44 weeks, approximately how many weeks have passed for your friend?

The perimeter, P, of a rectangle with length l and width w is given by the formula P=2l+2w. The area, A, is given by the formula A=lw. In Exercises 119120, use these formulas to find the perimeter and area of each rectangle. Express answers in simplified radical form. Remember that perimeter is measured in linear units, such as feet or meters, and area is measured in square units, such as square feet, ft2, or square meters, m2.

  1. 119.

    A rectangle of length radical 125 feet and width 2 radical 20 feet.
  2. 120.

    A rectangle of length 4 radical 20 feet and width radical 80 feet.

Explaining the Concepts

  1. 121. Explain how to simplify 105.

  2. 122. Explain how to add 3+12.

  3. 123. Describe what it means to rationalize a denominator. Use both 15 and 15+5 in your explanation.

  4. 124. What difference is there in simplifying (5)33 and (5)44?

  5. 125. What does amn mean?

  6. 126. Describe the kinds of numbers that have rational fifth roots.

  7. 127. Why must a and b represent nonnegative numbers when we write ab=ab? Is it necessary to use this restriction in the case of a3b3=ab3? Explain.

  8. 128. Read the Blitzer Bonus here. The future is now: You have the opportunity to explore the cosmos in a starship traveling near the speed of light. The experience will enable you to understand the mysteries of the universe in deeply personal ways, transporting you to unimagined levels of knowing and being. The downside: You return from your two-year journey to a futuristic world in which friends and loved ones are long gone. Do you explore space or stay here on Earth? What are the reasons for your choice?

Critical Thinking Exercises

Make Sense? In Exercises 129132, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 129. The unlike radicals 32 and 53 remind me of the unlike terms 3x and 5y that cannot be combined by addition or subtraction.

  2. 130. Using my calculator, I determined that 67=279,936, so 6 must be a seventh root of 279,936.

  3. 131. I simplified the terms of 220+475, and then I was able to add the like radicals.

  4. 132. When I use the definition for amn, I usually prefer to first raise a to the m power because smaller numbers are involved.

In Exercises 133136, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 133. 712712=49

  2. 134. 813=2

  3. 135. The cube root of 8 is not a real number.

  4. 136. 208=104

In Exercises 137138, fill in each box to make the statement true.

  1. 137. (5+00)(500)=22

  2. 138. 00x00=5x7

  3. 139. Find the exact value of 13+2+7 3+2 without the use of a calculator.

  4. 140. Place the correct symbol, > or <, in the shaded area between the given numbers. Do not use a calculator. Then check your result with a calculator.

    1. 312 00 313

    2. 7+18 00 7+18

  5. 141.

    1. A mathematics professor recently purchased a birthday cake for her son with the inscription

      Happy(252234÷214) th Birthday.

      How old is the son?

    2. The birthday boy, excited by the inscription on the cake, tried to wolf down the whole thing. Professor Mom, concerned about the possible metamorphosis of her son into a blimp, exclaimed, “Hold on! It is your birthday, so why not take 843+221634+21 of the cake? I’ll eat half of what’s left over.” How much of the cake did the professor eat?

Preview Exercises

Exercises 142144 will help you prepare for the material covered in the next section.

  1. 142. Multiply: (2x3 y2)(5x4 y7).

  2. 143. Use the distributive property to multiply:

    2x4(8x4+3x).
  3. 144. Simplify and express the answer in descending powers of x:

    2x(x2+4x+5)+3(x2+4x+5).
P.3: Exercise Set

P.3 Exercise Set

Practice Exercises

Evaluate each expression in Exercises 112, or indicate that the root is not a real number.

  1. 1. 36

  2. 2. 25

  3. 3. 36

  4. 4. 25

  5. 5. 36

  6. 6. 25

  7. 7. 2516

  8. 8. 144+25

  9. 9. 2516

  10. 10. 144+25

  11. 11. (13)2

  12. 12. (17)2

Use the product rule to simplify the expressions in Exercises 1322. In Exercises 1722, assume that variables represent nonnegative real numbers.

  1. 13. 50

  2. 14. 27

  3. 15. 45x2

  4. 16. 125x2

  5. 17. 2x6x

  6. 18. 10x 8x

  7. 19. x3

  8. 20. y3

  9. 21. 2x26x

  10. 22. 6x3x2

Use the quotient rule to simplify the expressions in Exercises 2332. Assume that x>0.

  1. 23. 181

  2. 24. 149

  3. 25. 4916

  4. 26. 1219

  5. 27. 48x33x

  6. 28. 72x38x

  7. 29. 150x43x

  8. 30. 24x43x

  9. 31. 200x310x1

  10. 32. 500x310x1

In Exercises 3344, add or subtract terms whenever possible.

  1. 33. 73+63

  2. 34. 85+115

  3. 35. 617x817x

  4. 36. 413x613x

  5. 37. 8+32

  6. 38. 20+65

  7. 39. 50x8x

  8. 40. 63x28x

  9. 41. 318+550

  10. 42. 412275

  11. 43. 3832+37275

  12. 44. 35422496+463

In Exercises 4554, rationalize the denominator.

  1. 45. 17

  2. 46. 210

  3. 47. 25

  4. 48. 73

  5. 49. 133+11

  6. 50. 33+7

  7. 51. 752

  8. 52. 531

  9. 53. 65+3

  10. 54. 1173

Evaluate each expression in Exercises 5566, or indicate that the root is not a real number.

  1. 55. 1253

  2. 56. 83

  3. 57. 83

  4. 58. 1253

  5. 59. 164

  6. 60. 814

  7. 61. (3) 44

  8. 62. (2) 44

  9. 63. (3)55

  10. 64. (2)55

  11. 65. 1325

  12. 66. 1646

Simplify the radical expressions in Exercises 6774, if possible.

  1. 67. 323

  2. 68. 1503

  3. 69. x43

  4. 70. x53

  5. 71. 9363

  6. 72. 12343

  7. 73. 64x652x5

  8. 74. 162x542x4

In Exercises 7582, add or subtract terms whenever possible.

  1. 75. 425+325

  2. 76. 635+235

  3. 77. 5163+543

  4. 78. 3243+813

  5. 79. 54xy33y128x3

  6. 80. 24xy33y81x3

  7. 81. 2+83

  8. 82. 3+153

In Exercises 8390, evaluate each expression without using a calculator.

  1. 83. 3612

  2. 84. 12112

  3. 85. 813

  4. 86. 2713

  5. 87. 12523

  6. 88. 823

  7. 89. 3245

  8. 90. 1652

In Exercises 91100, simplify using properties of exponents.

  1. 91. (7x13) (2x14)

  2. 92. (3x23) (4x34)

  3. 93. 20x125x14

  4. 94. 72x349x13

  5. 95. (x23)3

  6. 96. (x45)5

  7. 97. (25x4y6)12

  8. 98. (125x9y6)13

  9. 99. (3y14)3y112

  10. 100. (2y15)4y310

In Exercises 101108, simplify by reducing the index of the radical.

  1. 101. 524

  2. 102. 724

  3. 103. x63

  4. 104. x124

  5. 105. x46

  6. 106. x69

  7. 107. x6 y39

  8. 108. x4y812

Practice PLUS

In Exercises 109110, evaluate each expression.

  1. 109. 164+6253

  2. 110. 169+9+10003+21633

In Exercises 111114, simplify each expression. Assume that all variables represent positive numbers.

  1. 111. (49x2y4)12(xy12)

  2. 112. (8x6y3)13(x56y13)6

  3. 113. (x54y13x34)6

  4. 114. (x12y74y54)4

Application Exercises

  1. 115. Do you expect to pay more taxes than were withheld? Would you be surprised to know that the percentage of taxpayers who receive a refund and the percentage of taxpayers who pay more taxes vary according to age? The formula

    P=x(13+x)5x

    models the percentage, P, of taxpayers who are x years old who must pay more taxes.

    1. What percentage of 25-year-olds must pay more taxes?

    2. Rewrite the formula by rationalizing the denominator.

    3. Use the rationalized form of the formula from part (b) to find the percentage of 25-year-olds who must pay more taxes. Do you get the same answer as you did in part (a)? If so, does this prove that you correctly rationalized the denominator? Explain.

  2. 116. America is getting older. The graph shows the elderly U.S. population for ages 65–84 and for ages 85 and older in 2010, with projections for 2020 and beyond.

    A bar graph titled projected elderly United States population.

    Source: U.S. Census Bureau

    The formula E=5x+34.1 models the projected number of Americans ages 65–84, E, in millions, x years after 2010.

    1. Use the formula to find the projected increase in the number of Americans ages 65–84, in millions, from 2020 to 2050. Express this difference in simplified radical form.

    2. Use a calculator and write your answer in part (a) to the nearest tenth. Does this rounded decimal overestimate or underestimate the difference in the projected data shown by the bar graph? By how much?

  3. 117. The early Greeks believed that the most pleasing of all rectangles were golden rectangles, whose ratio of width to height is

    wh=251.

    The Parthenon at Athens fits into a golden rectangle once the triangular pediment is reconstructed.

    Rationalize the denominator of the given ratio, called the golden ratio. Then use a calculator and find the ratio of width to height, correct to the nearest hundredth, in golden rectangles.

  4. 118. Use Einstein’s special-relativity equation

    Ra=Rf1(vc)2,

    described in the Blitzer Bonus here, to solve this exercise. You are moving at 90% of the speed of light. Substitute 0.9c for v, your velocity, in the equation. What is your aging rate, correct to two decimal places, relative to a friend on Earth? If you are gone for 44 weeks, approximately how many weeks have passed for your friend?

The perimeter, P, of a rectangle with length l and width w is given by the formula P=2l+2w. The area, A, is given by the formula A=lw. In Exercises 119120, use these formulas to find the perimeter and area of each rectangle. Express answers in simplified radical form. Remember that perimeter is measured in linear units, such as feet or meters, and area is measured in square units, such as square feet, ft2, or square meters, m2.

  1. 119.

    A rectangle of length radical 125 feet and width 2 radical 20 feet.
  2. 120.

    A rectangle of length 4 radical 20 feet and width radical 80 feet.

Explaining the Concepts

  1. 121. Explain how to simplify 105.

  2. 122. Explain how to add 3+12.

  3. 123. Describe what it means to rationalize a denominator. Use both 15 and 15+5 in your explanation.

  4. 124. What difference is there in simplifying (5)33 and (5)44?

  5. 125. What does amn mean?

  6. 126. Describe the kinds of numbers that have rational fifth roots.

  7. 127. Why must a and b represent nonnegative numbers when we write ab=ab? Is it necessary to use this restriction in the case of a3b3=ab3? Explain.

  8. 128. Read the Blitzer Bonus here. The future is now: You have the opportunity to explore the cosmos in a starship traveling near the speed of light. The experience will enable you to understand the mysteries of the universe in deeply personal ways, transporting you to unimagined levels of knowing and being. The downside: You return from your two-year journey to a futuristic world in which friends and loved ones are long gone. Do you explore space or stay here on Earth? What are the reasons for your choice?

Critical Thinking Exercises

Make Sense? In Exercises 129132, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 129. The unlike radicals 32 and 53 remind me of the unlike terms 3x and 5y that cannot be combined by addition or subtraction.

  2. 130. Using my calculator, I determined that 67=279,936, so 6 must be a seventh root of 279,936.

  3. 131. I simplified the terms of 220+475, and then I was able to add the like radicals.

  4. 132. When I use the definition for amn, I usually prefer to first raise a to the m power because smaller numbers are involved.

In Exercises 133136, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 133. 712712=49

  2. 134. 813=2

  3. 135. The cube root of 8 is not a real number.

  4. 136. 208=104

In Exercises 137138, fill in each box to make the statement true.

  1. 137. (5+00)(500)=22

  2. 138. 00x00=5x7

  3. 139. Find the exact value of 13+2+7 3+2 without the use of a calculator.

  4. 140. Place the correct symbol, > or <, in the shaded area between the given numbers. Do not use a calculator. Then check your result with a calculator.

    1. 312 00 313

    2. 7+18 00 7+18

  5. 141.

    1. A mathematics professor recently purchased a birthday cake for her son with the inscription

      Happy(252234÷214) th Birthday.

      How old is the son?

    2. The birthday boy, excited by the inscription on the cake, tried to wolf down the whole thing. Professor Mom, concerned about the possible metamorphosis of her son into a blimp, exclaimed, “Hold on! It is your birthday, so why not take 843+221634+21 of the cake? I’ll eat half of what’s left over.” How much of the cake did the professor eat?

Preview Exercises

Exercises 142144 will help you prepare for the material covered in the next section.

  1. 142. Multiply: (2x3 y2)(5x4 y7).

  2. 143. Use the distributive property to multiply:

    2x4(8x4+3x).
  3. 144. Simplify and express the answer in descending powers of x:

    2x(x2+4x+5)+3(x2+4x+5).
Section P.4: Polynomials

Section P.4 Polynomials

Learning Objectives

What you’ll Learn

  1. 1 Understand the vocabulary of polynomials.

  2. 2 Add and subtract polynomials.

  3. 3 Multiply polynomials.

  4. 4 Use FOIL in polynomial multiplication.

  5. 5 Use special products in polynomial multiplication.

  6. 6 Perform operations with polynomials in several variables.

Old Dog ... New Chicks

Can that be Axl, your author’s yellow lab, sharing a special moment with a baby chick? And if it is (it is), what possible relevance can this have to polynomials? An answer is promised before you reach the Exercise Set. For now, we open the section by defining and describing polynomials.

How We Define Polynomials

More than 2 million people have tested their racial prejudice using an online version of the Implicit Association Test. Most groups’ average scores fall between “slight” and “moderate” bias, but the differences among age groups are intriguing.

In this section’s Exercise Set (Exercises 91 and 92), you will be working with a model that measures bias:

S=0.3x32.8x2+6.7x+30.

In this model, S represents the score on the Implicit Association Test. (Higher scores indicate stronger bias.) The variable x represents age group.

The algebraic expression that appears on the right side of the model is an example of a polynomial. A polynomial is a single term or the sum of two or more terms containing variables with whole-number exponents. This particular polynomial contains four terms. Equations containing polynomials are used in such diverse areas as science, business, medicine, psychology, and sociology. In this section, we review basic ideas about polynomials and their operations.

Objective 1: Understand the vocabulary of polynomials

How We Describe Polynomials

  1. Objective 1Understand the vocabulary of polynomials.

Watch Video

Consider the polynomial

7x39x2+13x6.

We can express 7x39x2+13x6 as

7x3+(9x2)+13x+(6).

The polynomial contains four terms. It is customary to write the terms in the order of descending powers of the variable. This is the standard form of a polynomial.

Some polynomials contain only one variable. Each term of such a polynomial in x is of the form axn. If a0, the degree of axn is n. For example, the degree of the term 7x3 is 3.

The Degree of axn

If a0, the degree of axn is n. The degree of a nonzero constant is 0. The constant 0 has no defined degree.

Here is an example of a polynomial and the degree of each of its four terms:

6 x to the fourth power labeled, degree 4, minus 3 x cubed labeled, degree 3, minus 2 x labeled, degree 1 minus 5 labeled, degree of nonzero constant, 0.

Notice that the exponent on x for the term 2x, meaning 2x1, is understood to be 1. For this reason, the degree of 2x is 1. You can think of 5 as 5x0; thus, its degree is 0.

A polynomial is simplified when it contains no grouping symbols and no like terms. A simplified polynomial that has exactly one term is called a monomial. A binomial is a simplified polynomial that has two terms. A trinomial is a simplified polynomial with three terms. Simplified polynomials with four or more terms have no special names.

The degree of a polynomial is the greatest degree of all the terms of the polynomial. For example, 4x2+3x is a binomial of degree 2 because the degree of the first term is 2, and the degree of the other term is less than 2. Also, 7x52x2+4 is a trinomial of degree 5 because the degree of the first term is 5, and the degrees of the other terms are less than 5.

Up to now, we have used x to represent the variable in a polynomial. However, any letter can be used. For example,

  • 7x53x3+8

is a polynomial (in x) of degree 5. Because there are three terms, the polynomial is a trinomial.
  • 6y3+4y2y+3

is a polynomial (in y) of degree 3. Because there are four terms, the polynomial has no special name.
  • z7+2

is a polynomial (in z) of degree 7. Because there are two terms, the polynomial is a binomial.

We can tie together the threads of our discussion with the formal definition of a polynomial in one variable. In this definition, the coefficients of the terms are represented by an  (read “a sub n”), an1 (read “a sub n minus 1”), an2, and so on. The small letters to the lower right of each a are called subscripts and are not exponents. Subscripts are used to distinguish one constant from another when a large and undetermined number of such constants are needed.

Definition of a Polynomial in x

A polynomial in x is an algebraic expression of the form

anxn+an1xn1+an2xn2++a1x+a0,

where an, an1, an2,, a1, and a0  are real numbers, an0, and n is a nonnegative integer. The polynomial is of degree n, an is the leading coefficient, and a0 is the constant term.

Objective 3: Multiply polynomials

Multiplying Polynomials

  1. Objective 3Multiply polynomials.

Watch Video

The product of two monomials is obtained by using properties of exponents. For example,

Left parenthesis negative 8 x to the sixth power right parenthesis left parenthesis 5 x to the third power = negative 8 times 5 x to the 6 + 3 power = negative 40 x to the ninth power. The text reads, multiply coefficients and add exponents.

Furthermore, we can use the distributive property to multiply a monomial and a polynomial that is not a monomial. For example,

Simplifying 3 x to the fourth power left parenthesis 2 x cubed minus 7 x + 3 right parenthesis.
P.4-83 Full Alternative Text

How do we multiply two polynomials if neither is a monomial? For example, consider

Left parenthesis 2 x + 3 right parenthesis left parenthesis x squared + 4 x + 5 right parenthesis. Left parenthesis 2 x + 3 right parenthesis is the binomial and left parenthesis x squared + 4 x + 5 right parenthesis is the trinomial.

One way to perform (2x+3)(x2+4x+5) is to distribute 2x throughout the trinomial

2x(x2+4x+5)

and 3 throughout the trinomial

3(x2+4x+5).

Then combine the like terms that result.

Multiplying Polynomials When Neither Is a Monomial

Multiply each term of one polynomial by each term of the other polynomial. Then combine like terms.

Example 2 Multiplying a Binomial and a Trinomial

Multiply: (2x+3)(x2+4x+5).

Solution

(2x+3)(x2+4x+5)=2x(x2+4x+5)+3(x2+4x+5)Multiply the trinomial by eachterm of the binomial.=2xx2+2x4x+2x5+3x2+34x+35Use the distributive property.=2x3+8x2+10x+3x2+12x+15Multiply monomials: Multiplycoefficients and add exponents.=2x3+11x2+22x+15Combine like terms:8x2+3x2=11x2 and10x+12x=22x.

Another method for performing the multiplication is to use a vertical format similar to that used for multiplying whole numbers.

The vertical format of multiplying two expressions.

Check Point 2

  • Multiply: (5x2)(3x25x+4).

Objective 4: Use FOIL in polynomial multiplication

The Product of Two Binomials: FOIL

  1. Objective 4Use FOIL in polynomial multiplication.

Watch Video

Frequently, we need to find the product of two binomials. One way to perform this multiplication is to distribute each term in the first binomial through the second binomial. For example, we can find the product of the binomials 3x+2 and 4x+5 as follows:

The process to simplify left parenthesis 3 x + 2 right parenthesis left parenthesis 4 x + 5 right parenthesis.
P.4-86 Full Alternative Text

We can also find the product of 3x+2 and 4x+5 using a method called FOIL, which is based on our preceding work. Any two binomials can be quickly multiplied by using the FOIL method, in which F represents the product of the first terms in each binomial, O represents the product of the outside terms, I represents the product of the inside terms, and L represents the product of the last, or second, terms in each binomial. For example, we can use the FOIL method to find the product of the binomials 3x+2 and 4x+5 as follows:

The process to solve the expression, left parenthesis 3 x + 2 right parenthesis left parenthesis 4 x + 5 right parenthesis, using the FOIL method.
P.4-87 Full Alternative Text

In general, here’s how to use the FOIL method to find the product of ax+b and cx+d:

Using the FOIL Method to Multiply Binomials

The process to solve the expression, left parenthesis a x + b right parenthesis left parenthesis c x + d right parenthesis, using the FOIL method.

Example 3 Using the FOIL Method

Multiply: (3x+4)(5x3).

Solution

The process to solve the expression, left parenthesis 3 x + 4 right parenthesis left parenthesis 5 x minus 3 right parenthesis, using the FOIL method.

Check Point 3

  • Multiply: (7x5)(4x3).

Objective 6: Perform operations with polynomials in several variables

Polynomials in Several Variables

  1. Objective 6Perform operations with polynomials in several variables.

Watch Video

A polynomial in two variables, x and y, contains the sum of one or more monomials in the form axn ym. The constant, a, is the coefficient. The exponents, n and m, represent whole numbers. The degree of the monomial axn ym is n+m.

Here is an example of a polynomial in two variables:

7 x squared y cubed minus 17 x to the fourth power y squared + x y minus 6y squared + 9. A note pointing at the numerals reads, the coefficients are 7, negative 17, 1, negative 6, and 9. The degrees of the monomials are also mentioned.
P.4-90 Full Alternative Text

The degree of a polynomial in two variables is the highest degree of all its terms. For the preceding polynomial, the degree is 6.

Polynomials containing two or more variables can be added, subtracted, and multiplied just like polynomials that contain only one variable. For example, we can add the monomials 7xy2 and 13xy2 as follows:

Negative 7 x y squared + 13 x y squared labeled, these like terms both contain the variable factors x and y squared.
P.4-91 Full Alternative Text

Example 4 Multiplying Polynomials in Two Variables

Multiply:

  1. (x+4y)(3x5y)

  2. (5x+3y)2.

Solution

We will perform the multiplication in part (a) using the FOIL method. We will multiply in part (b) using the formula for the square of a binomial sum, (A+B)2.

The Image shows the process to simplify a mathematical expression using the FOIL method.

Check Point 4

  • Multiply:

    1. (7x6y)(3xy)

    2. (2x+4y)2.

Special products can sometimes be used to find the products of certain trinomials, as illustrated in Example 5.

Example 5 Using the Special Products

Multiply:

  1. (7x+5+4y)(7x+54y)

  2. (3x+y+1)2.

Solution

  1. By grouping the first two terms within each of the parentheses, we can find the product using the form for the sum and difference of two terms.

    The formula, left parenthesis A + B right parenthesis times left parenthesis A minus B right parenthesis minus A squared minus B squared.
  2. We can group the terms of (3x+y+1)2 so that the formula for the square of a binomial can be applied.

    The formula, left parenthesis A + B right parenthesis squared = A squared + 2 times A times B + B squared.

Check Point 5

  • Multiply:

    1. (3x+2+5y)(3x+25 y)

    2. (2x+y+3)2.

P.4: Concept and Vocabulary Check

P.4 Concept and Vocabulary Check

Fill in each blank so that the resulting statement is true.

  1. C1. A polynomial is a single term or the sum of two or more terms containing variables with exponents that are ____ numbers.

  2. C2. It is customary to write the terms of a polynomial in the order of descending powers of the variable. This is called the ______ form of a polynomial.

  3. C3. A simplified polynomial that has exactly one term is called a/an ______.

  4. C4. A simplified polynomial that has two terms is called a/an ______.

  5. C5. A simplified polynomial that has three terms is called a/an ______.

  6. C6. If a0, the degree of axn is __.

  7. C7. Polynomials are added by combining _____ terms.

  8. C8. To multiply 7x3(4x58x2+6), use the _______ property to multiply each term of the trinomial _________ by the monomial _____.

  9. C9. To multiply (5x+3)(x2+8x+7), begin by multiplying each term of x2+8x+7 by _____. Then multiply each term of x2+8x+7 by ___. Then combine _____ terms.

  10. C10. When using the FOIL method to find (x+7)(3x+5), the product of the first terms is _____, the product of the outside terms is _____, the product of the inside terms is _____, and the product of the last terms is _____.

  11. C11. (A+B)(AB) = _______. The product of the sum and difference of the same two terms is the square of the first term ______ the square of the second term.

  12. C12. (A+B)2= _______. The square of a binomial sum is the first term ______ plus 2 times the ___________ plus the last term ______.

  13. C13. (AB)2= ______. The square of a binomial difference is the first term squared ______ 2 times the _______________ ______ the last term squared.

    Plus or minus?
  14. C14. If a0, the degree of axnym is _____.

P.4: Exercise Set

P.4 Exercise Set

Practice Exercises

In Exercises 14, is the algebraic expression a polynomial? If it is, write the polynomial in standard form.

  1. 1. 2x+3x25

  2. 2. 2x+3x15

  3. 3. 2x+3x

  4. 4. x2x3+x45

In Exercises 58, find the degree of the polynomial.

  1. 5. 3x25x+4

  2. 6. 4x3+7x211

  3. 7. x24x3+9x12x4+63

  4. 8. x28x3+15x4+91

In Exercises 914, perform the indicated operations. Write the resulting polynomial in standard form and indicate its degree.

  1. 9. (6x3+5x28x+9)+(17x3+2x24x13)

  2. 10. (7x3+6x211x+13)+(19x311x2+7x17)

  3. 11. (17x35x2+4x3)(5x39x28x+11)

  4. 12. (18x42x37x+8)(9x46x35x+7)

  5. 13. (5x27x8)+(2x23x+7)(x24x3)

  6. 14. (8x2+7x5)(3x24x)(6x35x2+3)

In Exercises 1582, find each product.

  1. 15. (x+1)(x2x+1)

  2. 16. (x+5)(x25x+25)

  3. 17. (2x3)(x23x+5)

  4. 18. (2x1)(x24x+3)

  5. 19. (x+7)(x+3)

  6. 20. (x+8)(x+5)

  7. 21. (x5)(x+3)

  8. 22. (x1)(x+2)

  9. 23. (3x+5)(2x+1)

  10. 24. (7x+4)(3x+1)

  11. 25. (2x3)(5x+3)

  12. 26. (2x5)(7x+2)

  13. 27. (5x24)(3x27)

  14. 28. (7x22)(3x25)

  15. 29. (8x3+3)(x25)

  16. 30. (7x3+5)(x22)

  17. 31. (x+3)(x3)

  18. 32. (x+5)(x5)

  19. 33. (3x+2)(3x2)

  20. 34. (2x+5)(2x5)

  21. 35. (57x)(5+7x)

  22. 36. (43x)(4+3x)

  23. 37. (4x2+5x)(4x25x)

  24. 38. (3x2+4x)(3x24x)

  25. 39. (1y5)(1+y5)

  26. 40. (2y5)(2+y5)

  27. 41. (x+2)2

  28. 42. (x+5)2

  29. 43. (2x+3)2

  30. 44. (3x+2)2

  31. 45. (x3)2

  32. 46. (x4)2

  33. 47. (4x21)2

  34. 48. (5x23)2

  35. 49. (72x)2

  36. 50. (95x)2

  37. 51. (x+1)3

  38. 52. (x+2)3

  39. 53. (2x+3)3

  40. 54. (3x+4)3

  41. 55. (x3)3

  42. 56. (x1)3

  43. 57. (3x4)3

  44. 58. (2x3)3

  45. 59. (x+5y)(7x+3y)

  46. 60. (x+9y)(6x+7y)

  47. 61. (x3y)(2x+7y)

  48. 62. (3xy)(2x+5y)

  49. 63. (3xy1)(5xy+2)

  50. 64. (7x2y+1)(2x2y3)

  51. 65. (7x+5y)2

  52. 66. (9x+7y)2

  53. 67. (x2y23)2

  54. 68. (x2y25)2

  55. 69. (xy)(x2+xy+y2)

  56. 70. (x+y)(x2xy+y2)

  57. 71. (3x+5y)(3x5y)

  58. 72. (7x+3y)(7x3y)

  59. 73. (x+y+3)(x+y3)

  60. 74. (x+y+5)(x+y5)

  61. 75. (3x+75y)(3x+7+5y)

  62. 76. (5x+7y2)(5x+7y+2)

  63. 77. [5y(2x+3)][5y+(2x+3)]

  64. 78. [8y+(73x)][8y(73x)]

  65. 79. (x+y+1)2

  66. 80. (x+y+2)2

  67. 81. (2x+y+1)2

  68. 82. (5x+1+6y)2

Practice PLUS

In Exercises 8390, perform the indicated operation or operations.

  1. 83. (3x+4y)2(3x4y)2

  2. 84. (5x+2y)2(5x2y)2

  3. 85. (5x7)(3x2)(4x5)(6x1)

  4. 86. (3x+5)(2x9)(7x2)(x1)

  5. 87. (2x+5)(2x5)(4x2+25)

  6. 88. (3x+4)(3x4)(9x2+16)

  7. 89. (2x7)5(2x7)3

  8. 90. (5x3)6(5x3)4

Application Exercises

The bar graph shows the differences among age groups on the Implicit Association Test that measures levels of racial prejudice. Higher scores indicate stronger bias. Exercises 9192 are based on the information in the graph. In these exercises, the possible values of the variable x are the group numbers shown in the voice balloons below the bars in the graph.

A bar graph titled measuring racial prejudice, by age.

Source: The Race Implicit Association Test on the Project Implicit Demonstration Website

  1. 91.

    1. The data can be described by the following polynomial model of degree 3:

      S=0.2x31.5x2+3.4x+25+(0.1x31.3x2+3.3x+5).

      In this polynomial model, S represents the score on the Implicit Association Test for age group x. Simplify the model.

    2. Use the simplified form of the model from part (a) to find the score on the Implicit Association Test for the group in the 45–54 age range. How well does the model describe the score displayed by the bar graph?

  2. 92.

    1. The data can be described by the following polynomial model of degree 3:

      S=0.1x31.2x2+2.4x+17+(0.2x31.6x2+4.3x+13).

      In this polynomial model, S represents the score on the Implicit Association Test for age group x. Simplify the model.

    2. Use the simplified form of the model from part (a) to find the score on the Implicit Association Test for the group in the 55–64 age range. How well does the model describe the score displayed by the bar graph?

The volume, V, of a rectangular solid with length l, width w, and height h is given by the formula V=lwh. In Exercises 9394, use this formula to write a polynomial in standard form that models, or represents, the volume of the open box.

  1. 93.

    An open box with a rectangular base had a length of 10 minus 2 x, width 8 minus 2 x, and depth x.
  2. 94.

    An open box with a rectangular base had a length 8 minus 2 x, width 5 minus 2 x, and depth x.

In Exercises 9596, write a polynomial in standard form that models, or represents, the area of the shaded region.

  1. 95.

    A large shaded rectangle of length x + 9 and width x + 3, has a smaller rectangle, of length x + 5 and width x + 1, within it. The area within the smaller rectangle is not shaded.
  2. 96.

    A large shaded rectangle of length x + 4 and width x + 3, has a smaller rectangle, of length x + 2 and width x + 1, within it. The area within the smaller rectangle is not shaded.

Explaining the Concepts

  1. 97. What is a polynomial in x?

  2. 98. Explain how to subtract polynomials.

  3. 99. Explain how to multiply two binomials using the FOIL method. Give an example with your explanation.

  4. 100. Explain how to find the product of the sum and difference of two terms. Give an example with your explanation.

  5. 101. Explain how to square a binomial difference. Give an example with your explanation.

  6. 102. Explain how to find the degree of a polynomial in two variables.

Critical Thinking Exercises

Make Sense? In Exercises 103106, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 103. Knowing the difference between factors and terms is important: In (3x2y)2, I can distribute the exponent 2 on each factor, but in (3x2+y)2, I cannot do the same thing on each term.

  2. 104. I used the FOIL method to find the product of x+5 and x2+2x+1.

  3. 105. Many English words have prefixes with meanings similar to those used to describe polynomials, such as monologue, binocular, and tricuspid.

  4. 106. Special-product formulas have patterns that make their multiplications quicker than using the FOIL method.

  5. 107. Express the area of the plane figure shown as a polynomial in standard form.

    A rectangle of length x + 3 and width x. The left side shows the width of x and the right side shows the width of x minus 1. The top right corner of the rectangle is removed and marked as x.

In Exercises 108109, represent the volume of each figure as a polynomial in standard form.

  1. 108.

    A cube of length x + 3 and width x, the left side shows height 2 x minus 1 and the right side shows height x + 1, and the top right corner of the part is removed is labeled x.
  2. 109.

    A figure shows rectangular cube hollow U-shaped has a length of 2 x + 1 and breath x + 5, the left side shows height x + 2. The inner part shows length x and the right side shows height 3.
  3. 110. Simplify: (yn+2)(yn2)(yn3)2.

Preview Exercises

Exercises 111113 will help you prepare for the material covered in the next section. In each exercise, replace the boxed question mark with an integer that results in the given product. Some trial and error may be necessary.

  1. 111. (x+3)(x+?)x2+7x+12

  2. 112. (x?)(x12)=x214x+24

  3. 113. (4x+1)(2x?)=8x210x3

P.4: Exercise Set

P.4 Exercise Set

Practice Exercises

In Exercises 14, is the algebraic expression a polynomial? If it is, write the polynomial in standard form.

  1. 1. 2x+3x25

  2. 2. 2x+3x15

  3. 3. 2x+3x

  4. 4. x2x3+x45

In Exercises 58, find the degree of the polynomial.

  1. 5. 3x25x+4

  2. 6. 4x3+7x211

  3. 7. x24x3+9x12x4+63

  4. 8. x28x3+15x4+91

In Exercises 914, perform the indicated operations. Write the resulting polynomial in standard form and indicate its degree.

  1. 9. (6x3+5x28x+9)+(17x3+2x24x13)

  2. 10. (7x3+6x211x+13)+(19x311x2+7x17)

  3. 11. (17x35x2+4x3)(5x39x28x+11)

  4. 12. (18x42x37x+8)(9x46x35x+7)

  5. 13. (5x27x8)+(2x23x+7)(x24x3)

  6. 14. (8x2+7x5)(3x24x)(6x35x2+3)

In Exercises 1582, find each product.

  1. 15. (x+1)(x2x+1)

  2. 16. (x+5)(x25x+25)

  3. 17. (2x3)(x23x+5)

  4. 18. (2x1)(x24x+3)

  5. 19. (x+7)(x+3)

  6. 20. (x+8)(x+5)

  7. 21. (x5)(x+3)

  8. 22. (x1)(x+2)

  9. 23. (3x+5)(2x+1)

  10. 24. (7x+4)(3x+1)

  11. 25. (2x3)(5x+3)

  12. 26. (2x5)(7x+2)

  13. 27. (5x24)(3x27)

  14. 28. (7x22)(3x25)

  15. 29. (8x3+3)(x25)

  16. 30. (7x3+5)(x22)

  17. 31. (x+3)(x3)

  18. 32. (x+5)(x5)

  19. 33. (3x+2)(3x2)

  20. 34. (2x+5)(2x5)

  21. 35. (57x)(5+7x)

  22. 36. (43x)(4+3x)

  23. 37. (4x2+5x)(4x25x)

  24. 38. (3x2+4x)(3x24x)

  25. 39. (1y5)(1+y5)

  26. 40. (2y5)(2+y5)

  27. 41. (x+2)2

  28. 42. (x+5)2

  29. 43. (2x+3)2

  30. 44. (3x+2)2

  31. 45. (x3)2

  32. 46. (x4)2

  33. 47. (4x21)2

  34. 48. (5x23)2

  35. 49. (72x)2

  36. 50. (95x)2

  37. 51. (x+1)3

  38. 52. (x+2)3

  39. 53. (2x+3)3

  40. 54. (3x+4)3

  41. 55. (x3)3

  42. 56. (x1)3

  43. 57. (3x4)3

  44. 58. (2x3)3

  45. 59. (x+5y)(7x+3y)

  46. 60. (x+9y)(6x+7y)

  47. 61. (x3y)(2x+7y)

  48. 62. (3xy)(2x+5y)

  49. 63. (3xy1)(5xy+2)

  50. 64. (7x2y+1)(2x2y3)

  51. 65. (7x+5y)2

  52. 66. (9x+7y)2

  53. 67. (x2y23)2

  54. 68. (x2y25)2

  55. 69. (xy)(x2+xy+y2)

  56. 70. (x+y)(x2xy+y2)

  57. 71. (3x+5y)(3x5y)

  58. 72. (7x+3y)(7x3y)

  59. 73. (x+y+3)(x+y3)

  60. 74. (x+y+5)(x+y5)

  61. 75. (3x+75y)(3x+7+5y)

  62. 76. (5x+7y2)(5x+7y+2)

  63. 77. [5y(2x+3)][5y+(2x+3)]

  64. 78. [8y+(73x)][8y(73x)]

  65. 79. (x+y+1)2

  66. 80. (x+y+2)2

  67. 81. (2x+y+1)2

  68. 82. (5x+1+6y)2

Practice PLUS

In Exercises 8390, perform the indicated operation or operations.

  1. 83. (3x+4y)2(3x4y)2

  2. 84. (5x+2y)2(5x2y)2

  3. 85. (5x7)(3x2)(4x5)(6x1)

  4. 86. (3x+5)(2x9)(7x2)(x1)

  5. 87. (2x+5)(2x5)(4x2+25)

  6. 88. (3x+4)(3x4)(9x2+16)

  7. 89. (2x7)5(2x7)3

  8. 90. (5x3)6(5x3)4

Application Exercises

The bar graph shows the differences among age groups on the Implicit Association Test that measures levels of racial prejudice. Higher scores indicate stronger bias. Exercises 9192 are based on the information in the graph. In these exercises, the possible values of the variable x are the group numbers shown in the voice balloons below the bars in the graph.

A bar graph titled measuring racial prejudice, by age.

Source: The Race Implicit Association Test on the Project Implicit Demonstration Website

  1. 91.

    1. The data can be described by the following polynomial model of degree 3:

      S=0.2x31.5x2+3.4x+25+(0.1x31.3x2+3.3x+5).

      In this polynomial model, S represents the score on the Implicit Association Test for age group x. Simplify the model.

    2. Use the simplified form of the model from part (a) to find the score on the Implicit Association Test for the group in the 45–54 age range. How well does the model describe the score displayed by the bar graph?

  2. 92.

    1. The data can be described by the following polynomial model of degree 3:

      S=0.1x31.2x2+2.4x+17+(0.2x31.6x2+4.3x+13).

      In this polynomial model, S represents the score on the Implicit Association Test for age group x. Simplify the model.

    2. Use the simplified form of the model from part (a) to find the score on the Implicit Association Test for the group in the 55–64 age range. How well does the model describe the score displayed by the bar graph?

The volume, V, of a rectangular solid with length l, width w, and height h is given by the formula V=lwh. In Exercises 9394, use this formula to write a polynomial in standard form that models, or represents, the volume of the open box.

  1. 93.

    An open box with a rectangular base had a length of 10 minus 2 x, width 8 minus 2 x, and depth x.
  2. 94.

    An open box with a rectangular base had a length 8 minus 2 x, width 5 minus 2 x, and depth x.

In Exercises 9596, write a polynomial in standard form that models, or represents, the area of the shaded region.

  1. 95.

    A large shaded rectangle of length x + 9 and width x + 3, has a smaller rectangle, of length x + 5 and width x + 1, within it. The area within the smaller rectangle is not shaded.
  2. 96.

    A large shaded rectangle of length x + 4 and width x + 3, has a smaller rectangle, of length x + 2 and width x + 1, within it. The area within the smaller rectangle is not shaded.

Explaining the Concepts

  1. 97. What is a polynomial in x?

  2. 98. Explain how to subtract polynomials.

  3. 99. Explain how to multiply two binomials using the FOIL method. Give an example with your explanation.

  4. 100. Explain how to find the product of the sum and difference of two terms. Give an example with your explanation.

  5. 101. Explain how to square a binomial difference. Give an example with your explanation.

  6. 102. Explain how to find the degree of a polynomial in two variables.

Critical Thinking Exercises

Make Sense? In Exercises 103106, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 103. Knowing the difference between factors and terms is important: In (3x2y)2, I can distribute the exponent 2 on each factor, but in (3x2+y)2, I cannot do the same thing on each term.

  2. 104. I used the FOIL method to find the product of x+5 and x2+2x+1.

  3. 105. Many English words have prefixes with meanings similar to those used to describe polynomials, such as monologue, binocular, and tricuspid.

  4. 106. Special-product formulas have patterns that make their multiplications quicker than using the FOIL method.

  5. 107. Express the area of the plane figure shown as a polynomial in standard form.

    A rectangle of length x + 3 and width x. The left side shows the width of x and the right side shows the width of x minus 1. The top right corner of the rectangle is removed and marked as x.

In Exercises 108109, represent the volume of each figure as a polynomial in standard form.

  1. 108.

    A cube of length x + 3 and width x, the left side shows height 2 x minus 1 and the right side shows height x + 1, and the top right corner of the part is removed is labeled x.
  2. 109.

    A figure shows rectangular cube hollow U-shaped has a length of 2 x + 1 and breath x + 5, the left side shows height x + 2. The inner part shows length x and the right side shows height 3.
  3. 110. Simplify: (yn+2)(yn2)(yn3)2.

Preview Exercises

Exercises 111113 will help you prepare for the material covered in the next section. In each exercise, replace the boxed question mark with an integer that results in the given product. Some trial and error may be necessary.

  1. 111. (x+3)(x+?)x2+7x+12

  2. 112. (x?)(x12)=x214x+24

  3. 113. (4x+1)(2x?)=8x210x3

Section P.5: Factoring Polynomials

Section P.5 Factoring Polynomials

Learning Objectives

What you’ll Learn

  1. 1 Factor out the greatest common factor of a polynomial.

  2. 2 Factor by grouping.

  3. 3 Factor trinomials.

  4. 4 Factor the difference of squares.

  5. 5 Factor perfect square trinomials.

  6. 6 Factor the sum or difference of two cubes.

  7. 7 Use a general strategy for factoring polynomials.

  8. 8 Factor algebraic expressions containing fractional and negative exponents.

A two-year-old boy is asked, “Do you have a brother?” He answers, “Yes.” “What is your brother’s name?” “Tom.” Asked if Tom has a brother, the two-year-old replies, “No.” The child can go in the direction from self to brother, but he cannot reverse this direction and move from brother back to self.

As our intellects develop, we learn to reverse the direction of our thinking. Reversibility of thought is found throughout algebra. For example, we can multiply polynomials and show that

5x(2x+3)=10x2+15x.

We can also reverse this process and express the resulting polynomial as

10x2+15x=5x(2x+3).

Factoring a polynomial expressed as the sum of monomials means finding an equivalent expression that is a product.

The process of factoring 10 x squared + 15 x. Sum of monomials, 10 x squared + 15 x, = an equivalent expression that is a product, 5 x left parenthesis 2 x + 3 right parenthesis. The factors of 10 x squared + 15 x are 5 x and 2 x + 3.

In this section, we will be factoring over the set of integers, meaning that the coefficients in the factors are integers. Polynomials that cannot be factored using integer coefficients are called irreducible over the integers, or prime.

The goal in factoring a polynomial is to use one or more factoring techniques until each of the polynomial’s factors, except possibly for a monomial factor, is prime or irreducible. In this situation, the polynomial is said to be factored completely.

We will now discuss basic techniques for factoring polynomials.

Objective 2: Factor by grouping

Factoring by Grouping

  1. Objective 2Factor by grouping.

Watch Video

Some polynomials have only a greatest common factor of 1. However, by a suitable grouping of the terms, it still may be possible to factor. This process, called factoring by grouping, is illustrated in Example 2.

Example 2 Factoring by Grouping

Factor: x3+4x2+3x+12.

Solution

There is no factor other than 1 common to all terms. However, we can group terms that have a common factor:

x cubed + 4 x squared + 3 x + 12. In x cubed + 4 x squared is labeled, the common factor is x squared. In 3 x + 12 is labeled, the common factor is 3.

We now factor the given polynomial as follows:

x3+4x2+3x+12=(x3+4x2)+(3x+12)Group terms with common factors.=x2(x+4)+3(x+4)Factor out the greatest common factorfrom the grouped terms. The remainingtwo terms have  x+4  as a commonbinomial factor.=(x+4)(x2+3).Factor out the GCF,  x+4

Thus, x3+4x2+3x+12=(x+4)(x2+3). Check the factorization by multiplying the right side of the equation using the FOIL method. Because the factorization is correct, you should obtain the original polynomial.

Check Point 2

  • Factor: x3+5x22x10.

Objective 3: Factor trinomials

Factoring Trinomials

  1. Objective 3Factor trinomials.

Watch Video

To factor a trinomial of the form ax2+bx+c, a little trial and error may be necessary.

A Strategy for Factoring ax2+bx+c

Assume, for the moment, that there is no greatest common factor.

  1. Find two First terms whose product is ax2:

    An illustration of how a x squared is obtained.
  2. Find two Last terms whose product is c:

    An illustration of how c is obtained.
  3. By trial and error, perform steps 1 and 2 until the sum of the Outside product and Inside product is bx:

    An illustration of how b x is obtained.

If no such combination exists, the polynomial is prime.

Example 3 Factoring a Trinomial Whose Leading Coefficient Is 1

Factor: x2+6x+8.

Solution

  1. Step 1 FIND TWO FIRST TERMS WHOSE PRODUCT IS x2.

    x2+6x+8=(x)(x)

  2. Step 1 (REPEATED) x2+6x+8=(x)(x)

  3. Step 2 FIND TWO LAST TERMS WHOSE PRODUCT IS 8.

    Factors of 8 8, 1 4, 2 8, 1 4, 2
  4. Step 3 TRY VARIOUS COMBINATIONS OF THESE FACTORS. The correct factorization of x2+6x+8 is the one in which the sum of the Outside and Inside products is equal to 6x. Here is a list of the possible factorizations:

    A table lists factorizations and their products.

Thus, x2+6x+8=(x+4)(x+2) or (x+2)(x+4).

In factoring a trinomial of the form x2+bx+c, you can speed things up by listing the factors of c and then finding their sums. We are interested in a sum of b. For example, in factoring x2+6x+8, we are interested in the factors of 8 whose sum is 6.

A table lists the factors of 8 and their sum.

Thus, x2+6x+8=(x+4)(x+2).

Check Point 3

  • Factor: x2+13x+40.

Example 4 Factoring a Trinomial Whose Leading Coefficient Is 1

Factor: x2+3x18.

Solution

  1. Step 1 FIND TWO FIRST TERMS WHOSE PRODUCT IS x2.

    x2+3x18=(x)(x)

    To find the second term of each factor, we must find two integers whose product is 18 and whose sum is 3.

  2. Step 2 FIND TWO LAST TERMS WHOSE PRODUCT IS 18.

    Factors of 18 18, 1 18, 1 9, 2 9, 2 6, 3 6, 3
  3. Step 3 TRY VARIOUS COMBINATIONS OF THESE FACTORS. We are looking for the pair of factors whose sum is 3.

    A table lists the factors of negative 18 and their sum.

    Thus, x2+3x18=(x+6)(x3) or (x3)(x+6).

Check Point 4

  • Factor: x25x14.

Example 5 Factoring a Trinomial Whose Leading Coefficient Is Not 1

Factor: 8x210x3.

Solution

  1. Step 1 FIND TWO FIRST TERMS WHOSE PRODUCT IS 8x2.

    8x210x3=?(8x)(x)  8x210x3=?(4x)(2x)
  2. Step 2 FIND TWO LAST TERMS WHOSE PRODUCT IS 3. The possible factorizations are 1(3) and 1(3).

  3. Step 3 TRY VARIOUS COMBINATIONS OF THESE FACTORS. The correct factorization of 8x210x3 is the one in which the sum of the Outside and Inside products is equal to 10x. Here is a list of the possible factorizations:

    A table lists factorizations and their products.

Thus, 8x210x3=(4x+1)(2x3)  or  (2x3)(4x+1).

Use FOIL multiplication to check either of these factorizations.

Check Point 5

  • Factor: 6x2+19x7.

Example 6 Factoring a Trinomial in Two Variables

Factor: 2x27xy+3y2.

Solution

  1. Step 1 FIND TWO FIRST TERMS WHOSE PRODUCT IS 2x2.

    2x27xy+3y2=(2x)(x)
  2. Step 2 FIND TWO LAST TERMS WHOSE PRODUCT IS 3y2. The possible factorizations are (y)(3y) and (y)(3y).

  3. Step 3 TRY VARIOUS COMBINATIONS OF THESE FACTORS. The correct factorization of 2x27xy+3y2 is the one in which the sum of the Outside and Inside products is equal to 7xy. Here is a list of possible factorizations:

    A table lists factorizations and their products.

Thus,

2x27xy+3y2=(2xy)(x3y)or(x3y)(2xy).

Use FOIL multiplication to check either of these factorizations.

Check Point 6

  • Factor: 3x213xy+4y2.

Objective 3: Factor trinomials

Factoring Trinomials

  1. Objective 3Factor trinomials.

Watch Video

To factor a trinomial of the form ax2+bx+c, a little trial and error may be necessary.

A Strategy for Factoring ax2+bx+c

Assume, for the moment, that there is no greatest common factor.

  1. Find two First terms whose product is ax2:

    An illustration of how a x squared is obtained.
  2. Find two Last terms whose product is c:

    An illustration of how c is obtained.
  3. By trial and error, perform steps 1 and 2 until the sum of the Outside product and Inside product is bx:

    An illustration of how b x is obtained.

If no such combination exists, the polynomial is prime.

Example 3 Factoring a Trinomial Whose Leading Coefficient Is 1

Factor: x2+6x+8.

Solution

  1. Step 1 FIND TWO FIRST TERMS WHOSE PRODUCT IS x2.

    x2+6x+8=(x)(x)

  2. Step 1 (REPEATED) x2+6x+8=(x)(x)

  3. Step 2 FIND TWO LAST TERMS WHOSE PRODUCT IS 8.

    Factors of 8 8, 1 4, 2 8, 1 4, 2
  4. Step 3 TRY VARIOUS COMBINATIONS OF THESE FACTORS. The correct factorization of x2+6x+8 is the one in which the sum of the Outside and Inside products is equal to 6x. Here is a list of the possible factorizations:

    A table lists factorizations and their products.

Thus, x2+6x+8=(x+4)(x+2) or (x+2)(x+4).

In factoring a trinomial of the form x2+bx+c, you can speed things up by listing the factors of c and then finding their sums. We are interested in a sum of b. For example, in factoring x2+6x+8, we are interested in the factors of 8 whose sum is 6.

A table lists the factors of 8 and their sum.

Thus, x2+6x+8=(x+4)(x+2).

Check Point 3

  • Factor: x2+13x+40.

Example 4 Factoring a Trinomial Whose Leading Coefficient Is 1

Factor: x2+3x18.

Solution

  1. Step 1 FIND TWO FIRST TERMS WHOSE PRODUCT IS x2.

    x2+3x18=(x)(x)

    To find the second term of each factor, we must find two integers whose product is 18 and whose sum is 3.

  2. Step 2 FIND TWO LAST TERMS WHOSE PRODUCT IS 18.

    Factors of 18 18, 1 18, 1 9, 2 9, 2 6, 3 6, 3
  3. Step 3 TRY VARIOUS COMBINATIONS OF THESE FACTORS. We are looking for the pair of factors whose sum is 3.

    A table lists the factors of negative 18 and their sum.

    Thus, x2+3x18=(x+6)(x3) or (x3)(x+6).

Check Point 4

  • Factor: x25x14.

Example 5 Factoring a Trinomial Whose Leading Coefficient Is Not 1

Factor: 8x210x3.

Solution

  1. Step 1 FIND TWO FIRST TERMS WHOSE PRODUCT IS 8x2.

    8x210x3=?(8x)(x)  8x210x3=?(4x)(2x)
  2. Step 2 FIND TWO LAST TERMS WHOSE PRODUCT IS 3. The possible factorizations are 1(3) and 1(3).

  3. Step 3 TRY VARIOUS COMBINATIONS OF THESE FACTORS. The correct factorization of 8x210x3 is the one in which the sum of the Outside and Inside products is equal to 10x. Here is a list of the possible factorizations:

    A table lists factorizations and their products.

Thus, 8x210x3=(4x+1)(2x3)  or  (2x3)(4x+1).

Use FOIL multiplication to check either of these factorizations.

Check Point 5

  • Factor: 6x2+19x7.

Example 6 Factoring a Trinomial in Two Variables

Factor: 2x27xy+3y2.

Solution

  1. Step 1 FIND TWO FIRST TERMS WHOSE PRODUCT IS 2x2.

    2x27xy+3y2=(2x)(x)
  2. Step 2 FIND TWO LAST TERMS WHOSE PRODUCT IS 3y2. The possible factorizations are (y)(3y) and (y)(3y).

  3. Step 3 TRY VARIOUS COMBINATIONS OF THESE FACTORS. The correct factorization of 2x27xy+3y2 is the one in which the sum of the Outside and Inside products is equal to 7xy. Here is a list of possible factorizations:

    A table lists factorizations and their products.

Thus,

2x27xy+3y2=(2xy)(x3y)or(x3y)(2xy).

Use FOIL multiplication to check either of these factorizations.

Check Point 6

  • Factor: 3x213xy+4y2.

Objective 3: Factor trinomials

Factoring Trinomials

  1. Objective 3Factor trinomials.

Watch Video

To factor a trinomial of the form ax2+bx+c, a little trial and error may be necessary.

A Strategy for Factoring ax2+bx+c

Assume, for the moment, that there is no greatest common factor.

  1. Find two First terms whose product is ax2:

    An illustration of how a x squared is obtained.
  2. Find two Last terms whose product is c:

    An illustration of how c is obtained.
  3. By trial and error, perform steps 1 and 2 until the sum of the Outside product and Inside product is bx:

    An illustration of how b x is obtained.

If no such combination exists, the polynomial is prime.

Example 3 Factoring a Trinomial Whose Leading Coefficient Is 1

Factor: x2+6x+8.

Solution

  1. Step 1 FIND TWO FIRST TERMS WHOSE PRODUCT IS x2.

    x2+6x+8=(x)(x)

  2. Step 1 (REPEATED) x2+6x+8=(x)(x)

  3. Step 2 FIND TWO LAST TERMS WHOSE PRODUCT IS 8.

    Factors of 8 8, 1 4, 2 8, 1 4, 2
  4. Step 3 TRY VARIOUS COMBINATIONS OF THESE FACTORS. The correct factorization of x2+6x+8 is the one in which the sum of the Outside and Inside products is equal to 6x. Here is a list of the possible factorizations:

    A table lists factorizations and their products.

Thus, x2+6x+8=(x+4)(x+2) or (x+2)(x+4).

In factoring a trinomial of the form x2+bx+c, you can speed things up by listing the factors of c and then finding their sums. We are interested in a sum of b. For example, in factoring x2+6x+8, we are interested in the factors of 8 whose sum is 6.

A table lists the factors of 8 and their sum.

Thus, x2+6x+8=(x+4)(x+2).

Check Point 3

  • Factor: x2+13x+40.

Example 4 Factoring a Trinomial Whose Leading Coefficient Is 1

Factor: x2+3x18.

Solution

  1. Step 1 FIND TWO FIRST TERMS WHOSE PRODUCT IS x2.

    x2+3x18=(x)(x)

    To find the second term of each factor, we must find two integers whose product is 18 and whose sum is 3.

  2. Step 2 FIND TWO LAST TERMS WHOSE PRODUCT IS 18.

    Factors of 18 18, 1 18, 1 9, 2 9, 2 6, 3 6, 3
  3. Step 3 TRY VARIOUS COMBINATIONS OF THESE FACTORS. We are looking for the pair of factors whose sum is 3.

    A table lists the factors of negative 18 and their sum.

    Thus, x2+3x18=(x+6)(x3) or (x3)(x+6).

Check Point 4

  • Factor: x25x14.

Example 5 Factoring a Trinomial Whose Leading Coefficient Is Not 1

Factor: 8x210x3.

Solution

  1. Step 1 FIND TWO FIRST TERMS WHOSE PRODUCT IS 8x2.

    8x210x3=?(8x)(x)  8x210x3=?(4x)(2x)
  2. Step 2 FIND TWO LAST TERMS WHOSE PRODUCT IS 3. The possible factorizations are 1(3) and 1(3).

  3. Step 3 TRY VARIOUS COMBINATIONS OF THESE FACTORS. The correct factorization of 8x210x3 is the one in which the sum of the Outside and Inside products is equal to 10x. Here is a list of the possible factorizations:

    A table lists factorizations and their products.

Thus, 8x210x3=(4x+1)(2x3)  or  (2x3)(4x+1).

Use FOIL multiplication to check either of these factorizations.

Check Point 5

  • Factor: 6x2+19x7.

Example 6 Factoring a Trinomial in Two Variables

Factor: 2x27xy+3y2.

Solution

  1. Step 1 FIND TWO FIRST TERMS WHOSE PRODUCT IS 2x2.

    2x27xy+3y2=(2x)(x)
  2. Step 2 FIND TWO LAST TERMS WHOSE PRODUCT IS 3y2. The possible factorizations are (y)(3y) and (y)(3y).

  3. Step 3 TRY VARIOUS COMBINATIONS OF THESE FACTORS. The correct factorization of 2x27xy+3y2 is the one in which the sum of the Outside and Inside products is equal to 7xy. Here is a list of possible factorizations:

    A table lists factorizations and their products.

Thus,

2x27xy+3y2=(2xy)(x3y)or(x3y)(2xy).

Use FOIL multiplication to check either of these factorizations.

Check Point 6

  • Factor: 3x213xy+4y2.

Objective 4: Factor the difference of squares

Factoring the Difference of Two Squares

  1. Objective 4Factor the difference of squares.

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A method for factoring the difference of two squares is obtained by reversing the special product for the sum and difference of two terms.

The Difference of Two Squares

If A and B are real numbers, variables, or algebraic expressions, then

A2B2=(A+B)(AB).

In words: The difference of the squares of two terms factors as the product of a sum and a difference of those terms.

Example 7 Factoring the Difference of Two Squares

Factor:

  1. x24

  2. 81x249.

Solution

We must express each term as the square of some monomial. Then we use the formula for factoring A2B2.

The formula A squared minus B squared = left parenthesis A + B right parenthesis, left parenthesis A minus B right parenthesis, is applied to simplify the equations.

Check Point 7

  • Factor:

    1. x281

    2. 36x225.

We have seen that a polynomial is factored completely when it is written as the product of prime polynomials. To be sure that you have factored completely, check to see whether any factors with more than one term in the factored polynomial can be factored further. If so, continue factoring.

Example 8 A Repeated Factorization

Factor completely: x481.

Solution

x481=(x2)292Express as the difference of two squares.=(x2+9)(x29)The factors are the sum and the difference of theexpressions being squared.=(x2+9)(x232)The factor  x29  is the difference of two squaresand can be factored.=(x2+9)(x+3)(x3)The factors of  x29 are the sum and thedifference of the expressions being squared.

Check Point 8

  • Factor completely: 81x416.

Objective 5: Factor perfect square trinomials

Factoring Perfect Square Trinomials

  1. Objective 5Factor perfect square trinomials.

Watch Video

Our next factoring technique is obtained by reversing the special products for squaring binomials. The trinomials that are factored using this technique are called perfect square trinomials.

Factoring Perfect Square Trinomials

Let A and B be real numbers, variables, or algebraic expressions.

The process of factoring perfect square trinomials.

The two items in the box show that perfect square trinomials, A2+2AB+B2 and A22AB+B2, come in two forms: one in which the coefficient of the middle term is positive and one in which the coefficient of the middle term is negative. Here’s how to recognize a perfect square trinomial:

  1. The first and last terms are squares of monomials or integers.

  2. The middle term is twice the product of the expressions being squared in the first and last terms.

Example 9 Factoring Perfect Square Trinomials

Factor:

  1. x2+6x+9

  2. 25x260x+36.

Solution

  1.  

    The formula, A squared + 2 A B + B squared = left parenthesis A + B right parenthesis squared is applied to solve the equation.

  2. We suspect that 25x260x+36 is a perfect square trinomial because 25x2=(5x)2 and 36=62. The middle term can be expressed as twice the product of 5x and 6.

    The process to solve a trinomial.

Check Point 9

  • Factor:

    1. x2+14x+49

    2. 16x256x+49.

Objective 7: Use a general strategy for factoring polynomials

A Strategy for Factoring Polynomials

  1. Objective 7Use a general strategy for factoring polynomials.

Watch Video

It is important to practice factoring a wide variety of polynomials so that you can quickly select the appropriate technique. The polynomial is factored completely when all its polynomial factors, except possibly for monomial factors, are prime. Because of the commutative property, the order of the factors does not matter.

A Strategy for Factoring a Polynomial

  1. If there is a common factor, factor out the GCF.

  2. Determine the number of terms in the polynomial and try factoring as follows:

    1. If there are two terms, can the binomial be factored by using one of the following special forms?

      Difference of two squares:A2B2=(A+B)(AB)Sum of two cubes:A3+B3=(A+B)(A2AB+B2)Difference of two cubes: A3B3=(AB)(A2+AB+B2)
    2. If there are three terms, is the trinomial a perfect square trinomial? If so, factor by using one of the following special forms:

      A2+2AB+B2=(A+B)2A22AB+B2=(AB)2.

      If the trinomial is not a perfect square trinomial, try factoring by trial and error.

    3. If there are four or more terms, try factoring by grouping.

  3. Check to see if any factors with more than one term in the factored polynomial can be factored further. If so, factor completely.

Example 11 Factoring a Polynomial

Factor: 2x3+8x2+8x.

Solution

  1. Step 1 IF THERE IS A COMMON FACTOR, FACTOR OUT THE GCF. Because 2x is common to all terms, we factor it out.

    2x3+8x2+8x=2x(x2+4x+4)Factor out the GCF.
  2. Step 2 DETERMINE THE NUMBER OF TERMS AND FACTOR ACCORDINGLY. The factor x2+4x+4 has three terms and is a perfect square trinomial. We factor using A2+2AB+B2=(A+B)2.

    The process to simplify an equation.
  3. Step 3 CHECK TO SEE IF FACTORS CAN BE FACTORED FURTHER. In this problem, they cannot. Thus,

    2x3+8x2+8x=2x(x+2)2.

Check Point 11

  • Factor: 3x330x2+75x.

Example 12 Factoring a Polynomial

Factor: x225a2+8x+16.

Solution

  1. Step 1 IF THERE IS A COMMON FACTOR, FACTOR OUT THE GCF. Other than 1 or 1, there is no common factor.

  2. Step 2 DETERMINE THE NUMBER OF TERMS AND FACTOR ACCORDINGLY. There are four terms. We try factoring by grouping. It can be shown that grouping into two groups of two terms does not result in a common binomial factor. Let’s try grouping as a difference of squares.

    x225a2+8x+16=(x2+8x+16)25a2Rearrange terms and group as a perfect squaretrinomial minus  25a2  to obtain a difference ofsquares.=(x+4)2(5a)2Factor the perfect square trinomial.=(x+4+5a)(x+45a)Factor the difference of squares. Thefactors are the sum and difference of theexpressions being squared.
  3. Step 3 CHECK TO SEE IF FACTORS CAN BE FACTORED FURTHER. In this case, they cannot, so we have factored completely.

Check Point 12

  • Factor: x236a2+20x+100.

Objective 7: Use a general strategy for factoring polynomials

A Strategy for Factoring Polynomials

  1. Objective 7Use a general strategy for factoring polynomials.

Watch Video

It is important to practice factoring a wide variety of polynomials so that you can quickly select the appropriate technique. The polynomial is factored completely when all its polynomial factors, except possibly for monomial factors, are prime. Because of the commutative property, the order of the factors does not matter.

A Strategy for Factoring a Polynomial

  1. If there is a common factor, factor out the GCF.

  2. Determine the number of terms in the polynomial and try factoring as follows:

    1. If there are two terms, can the binomial be factored by using one of the following special forms?

      Difference of two squares:A2B2=(A+B)(AB)Sum of two cubes:A3+B3=(A+B)(A2AB+B2)Difference of two cubes: A3B3=(AB)(A2+AB+B2)
    2. If there are three terms, is the trinomial a perfect square trinomial? If so, factor by using one of the following special forms:

      A2+2AB+B2=(A+B)2A22AB+B2=(AB)2.

      If the trinomial is not a perfect square trinomial, try factoring by trial and error.

    3. If there are four or more terms, try factoring by grouping.

  3. Check to see if any factors with more than one term in the factored polynomial can be factored further. If so, factor completely.

Example 11 Factoring a Polynomial

Factor: 2x3+8x2+8x.

Solution

  1. Step 1 IF THERE IS A COMMON FACTOR, FACTOR OUT THE GCF. Because 2x is common to all terms, we factor it out.

    2x3+8x2+8x=2x(x2+4x+4)Factor out the GCF.
  2. Step 2 DETERMINE THE NUMBER OF TERMS AND FACTOR ACCORDINGLY. The factor x2+4x+4 has three terms and is a perfect square trinomial. We factor using A2+2AB+B2=(A+B)2.

    The process to simplify an equation.
  3. Step 3 CHECK TO SEE IF FACTORS CAN BE FACTORED FURTHER. In this problem, they cannot. Thus,

    2x3+8x2+8x=2x(x+2)2.

Check Point 11

  • Factor: 3x330x2+75x.

Example 12 Factoring a Polynomial

Factor: x225a2+8x+16.

Solution

  1. Step 1 IF THERE IS A COMMON FACTOR, FACTOR OUT THE GCF. Other than 1 or 1, there is no common factor.

  2. Step 2 DETERMINE THE NUMBER OF TERMS AND FACTOR ACCORDINGLY. There are four terms. We try factoring by grouping. It can be shown that grouping into two groups of two terms does not result in a common binomial factor. Let’s try grouping as a difference of squares.

    x225a2+8x+16=(x2+8x+16)25a2Rearrange terms and group as a perfect squaretrinomial minus  25a2  to obtain a difference ofsquares.=(x+4)2(5a)2Factor the perfect square trinomial.=(x+4+5a)(x+45a)Factor the difference of squares. Thefactors are the sum and difference of theexpressions being squared.
  3. Step 3 CHECK TO SEE IF FACTORS CAN BE FACTORED FURTHER. In this case, they cannot, so we have factored completely.

Check Point 12

  • Factor: x236a2+20x+100.

Objective 8: Factor algebraic expressions containing fractional and negative exponents

Factoring Algebraic Expressions Containing Fractional and Negative Exponents

  1. Objective 8Factor algebraic expressions containing fractional and negative exponents.

Watch Video

Although expressions containing fractional and negative exponents are not polynomials, they can be simplified using factoring techniques.

Example 13 Factoring Involving Fractional and Negative Exponents

Factor and simplify: x(x+1)34+(x+1)14.

Solution

The greatest common factor of x(x+1)34+(x+1)14 is x+1 with the smaller exponent in the two terms. Thus, the greatest common factor is (x+1)34.

x(x+1)34+(x+1)14=(x+1)34x+(x+1)34(x+1)44Express each term as the productof the greatest common factorand its other factor. Note that(x+1)34(x+1)44=(x+1)34+44=(x+1)14.=(x+1)34x+(x+1)34(x+1)Simplify:  (x+1)44=(x+1).=(x+1)34[x+(x+1)]Factor out the greatest common factor.=2x+1(x+1)34bn=1bn

Check Point 13

  • Factor and simplify: x(x1)12+(x1)12.

P.5: Exercise Set

P.5 Exercise Set

Practice Exercises

In Exercises 110, factor out the greatest common factor.

  1. 1. 18x+27

  2. 2. 16x24

  3. 3. 3x2+6x

  4. 4. 4x28x

  5. 5. 9x418x3+27x2

  6. 6. 6x418x3+12x2

  7. 7. x(x+5)+3(x+5)

  8. 8. x(2x+1)+4(2x+1)

  9. 9. x2(x3)+12(x3)

  10. 10. x2(2x+5)+17(2x+5)

In Exercises 1116, factor by grouping.

  1. 11. x32x2+5x10

  2. 12. x33x2+4x12

  3. 13. x3x2+2x2

  4. 14. x3+6x22x12

  5. 15. 3x32x26x+4

  6. 16. x3x25x+5

In Exercises 1738, factor each trinomial, or state that the trinomial is prime.

  1. 17. x2+5x+6

  2. 18. x2+8x+15

  3. 19. x22x15

  4. 20. x24x5

  5. 21. x28x+15

  6. 22. x214x+45

  7. 23. 3x2x2

  8. 24. 2x2+5x3

  9. 25. 3x225x28

  10. 26. 3x22x5

  11. 27. 6x211x+4

  12. 28. 6x217x+12

  13. 29. 4x2+16x+15

  14. 30. 8x2+33x+4

  15. 31. 9x29x+2

  16. 32. 9x2+5x4

  17. 33. 20x2+27x8

  18. 34. 15x219x+6

  19. 35. 2x2+3xy+y2

  20. 36. 3x2+4xy+y2

  21. 37. 6x25xy6y2

  22. 38. 6x27xy5y2

In Exercises 3948, factor the difference of two squares.

  1. 39. x2100

  2. 40. x2144

  3. 41. 36x249

  4. 42. 64x281

  5. 43. 9x225y2

  6. 44. 36x249y2

  7. 45. x416

  8. 46. x41

  9. 47. 16x481

  10. 48. 81x41

In Exercises 4956, factor each perfect square trinomial.

  1. 49. x2+2x+1

  2. 50. x2+4x+4

  3. 51. x214x+49

  4. 52. x210x+25

  5. 53. 4x2+4x+1

  6. 54. 25x2+10x+1

  7. 55. 9x26x+1

  8. 56. 64x216x+1

In Exercises 5764, factor using the formula for the sum or difference of two cubes.

  1. 57. x3+27

  2. 58. x3+64

  3. 59. x364

  4. 60. x327

  5. 61. 8x31

  6. 62. 27x31

  7. 63. 64x3+27

  8. 64. 8x3+125

In Exercises 6592, factor completely, or state that the polynomial is prime.

  1. 65. 3x33x

  2. 66. 5x345x

  3. 67. 4x24x24

  4. 68. 6x218x60

  5. 69. 2x4162

  6. 70. 7x47

  7. 71. x3+2x29x18

  8. 72. x3+3x225x75

  9. 73. 2x22x112

  10. 74. 6x26x12

  11. 75. x34x

  12. 76. 9x39x

  13. 77. x2+64

  14. 78. x2+36

  15. 79. x3+2x24x8

  16. 80. x3+2x2x2

  17. 81. y581y

  18. 82. y516y

  19. 83. 20y445y2

  20. 84. 48y43y2

  21. 85. x212x+3649y2

  22. 86. x210x+2536y2

  23. 87. 9b2 x16y16x+9b2 y

  24. 88. 16a2 x25y25x+16a2 y

  25. 89. x2 y16y+322x2

  26. 90. 12x2 y27y4x2+9

  27. 91. 2x38a2 x+24x2+72x

  28. 92. 2x398a2 x+28x2+98x

In Exercises 93102, factor and simplify each algebraic expression.

  1. 93. x32x12

  2. 94. x34x14

  3. 95. 4x23+8x13

  4. 96. 12x34+6x14

  5. 97. (x+3)12(x+3)32

  6. 98. (x2+4)32+(x2+4)72

  7. 99. (x+5)12(x+5)32

  8. 100. (x2+3)23+(x2+3)53

  9. 101. (4x1)1213(4x1)32

  10. 102. 8(4x+3)2+10(5x+1)(4x+3)1

Practice PLUS

In Exercises 103114, factor completely.

  1. 103. 10x2(x+1)7x(x+1)6(x+1)

  2. 104. 12x2(x1)4x(x1)5(x1)

  3. 105. 6x4+35x26

  4. 106. 7x4+34x25

  5. 107. y7+y

  6. 108. (y+1)3+1

  7. 109. x45x2 y2+4y4

  8. 110. x410x2 y2+9y4

  9. 111. (xy)44(xy)2

  10. 112. (x+y)4100(x+y)2

  11. 113. 2x27xy2+3y4

  12. 114. 3x2+5xy2+2y4

Application Exercises

  1. 115. Your electronics store is having an incredible sale. The price on one laptop is reduced by 40%. Then the sale price is reduced by another 40%. If x is the laptop’s original price, the sale price can be modeled by

    (x0.4x)0.4(x0.4x).
    1. Factor out (x0.4x) from each term. Then simplify the resulting expression.

    2. Use the simplified expression from part (a) to answer these questions. With a 40% reduction followed by a 40% reduction, is the laptop selling at 20% of its original price? If not, at what percentage of the original price is it selling?

  2. 116. Your local electronics store is having an end-of-the-year sale. The price on an Ultra 4K HD television had been reduced by 30%. Now the sale price is reduced by another 30%. If x is the television’s original price, the sale price can be modeled by

    (x0.3x)0.3(x0.3x).
    1. Factor out (x0.3x) from each term. Then simplify the resulting expression.

    2. Use the simplified expression from part (a) to answer these questions. With a 30% reduction followed by a 30% reduction, is the television selling at 40% of its original price? If not, at what percentage of the original price is it selling?

In Exercises 117120,

  1. Write an expression for the area of the shaded region.

  2. Write the expression in factored form.

  1. 117.

    A large squared with sides of length 3 x is divided into 4 smaller squares with sides of length 2, and a plus shaped the central shaded region.
  2. 118.

    A large square with sides of length 7 x is divided into 4 smaller squares at each corner with sides of length 3, and a plus shaped central shaded region within the first square but outside the small squares.
  3. 119.

    A rectangle of length x + y and width x. A smaller rectangle of length x + y and width y is drawn within the first rectangle along its center. The region within the first rectangle but outside the second rectangle is shaded.
  4. 120.

    A diagram is constructed with two squares and a rectangle.

In Exercises 121122, find the formula for the volume of the region outside the smaller rectangular solid and inside the larger rectangular solid. Then express the volume in factored form.

  1. 121.

    A rectangular solid of width b and height b is positioned along the length, at the center, of a larger rectangular solid of length 4 a, width a, and height a.
  2. 122.

    A rectangular solid of width b and height b is positioned along the length, at the center, of a larger rectangular solid of length 3 a, width a, and height a.

Explaining the Concepts

  1. 123. Using an example, explain how to factor out the greatest common factor of a polynomial.

  2. 124. Suppose that a polynomial contains four terms. Explain how to use factoring by grouping to factor the polynomial.

  3. 125. Explain how to factor 3x2+10x+8.

  4. 126. Explain how to factor the difference of two squares. Provide an example with your explanation.

  5. 127. What is a perfect square trinomial and how is it factored?

  6. 128. Explain how to factor x3+1.

  7. 129. What does it mean to factor completely?

Critical Thinking Exercises

Make Sense? In Exercises 130133, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 130. Although 20x3 appears in both 20x3+8x2 and 20x3+10x, I’ll need to factor 20x3 in different ways to obtain each polynomial’s factorization.

  2. 131. You grouped the polynomial’s terms using different groupings than I did, yet we both obtained the same factorization.

  3. 132. I factored 4x2100 completely and obtained (2x+10) (2x10).

  4. 133. First factoring out the greatest common factor makes it easier for me to determine how to factor the remaining factor, assuming that it is not prime.

In Exercises 134137, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 134. x416 is factored completely as (x2+4)(x24).

  2. 135. The trinomial x24x4 is a prime polynomial.

  3. 136. x2+36=(x+6)2

  4. 137. x364=(x+4)(x2+4x16)

In Exercises 138141, factor completely.

  1. 138. x2n+6xn+8

  2. 139. x24x+5

  3. 140. x4y42x3 y+2xy3

  4. 141. (x5)12(x+5)12(x+5)12(x5)32

In Exercises 142143, find all integers b so that the trinomial can be factored.

  1. 142. x2+bx+15

  2. 143. x2+4x+b

Preview Exercises

Exercises 144146 will help you prepare for the material covered in the next section.

  1. 144. Factor the numerator and the denominator. Then simplify by dividing out the common factor in the numerator and the denominator.

    x2+6x+5x225

In Exercises 145146, perform the indicated operation. Where possible, reduce the answer to its lowest terms.

  1. 145. 54815

  2. 146. 12+23

P.5: Exercise Set

P.5 Exercise Set

Practice Exercises

In Exercises 110, factor out the greatest common factor.

  1. 1. 18x+27

  2. 2. 16x24

  3. 3. 3x2+6x

  4. 4. 4x28x

  5. 5. 9x418x3+27x2

  6. 6. 6x418x3+12x2

  7. 7. x(x+5)+3(x+5)

  8. 8. x(2x+1)+4(2x+1)

  9. 9. x2(x3)+12(x3)

  10. 10. x2(2x+5)+17(2x+5)

In Exercises 1116, factor by grouping.

  1. 11. x32x2+5x10

  2. 12. x33x2+4x12

  3. 13. x3x2+2x2

  4. 14. x3+6x22x12

  5. 15. 3x32x26x+4

  6. 16. x3x25x+5

In Exercises 1738, factor each trinomial, or state that the trinomial is prime.

  1. 17. x2+5x+6

  2. 18. x2+8x+15

  3. 19. x22x15

  4. 20. x24x5

  5. 21. x28x+15

  6. 22. x214x+45

  7. 23. 3x2x2

  8. 24. 2x2+5x3

  9. 25. 3x225x28

  10. 26. 3x22x5

  11. 27. 6x211x+4

  12. 28. 6x217x+12

  13. 29. 4x2+16x+15

  14. 30. 8x2+33x+4

  15. 31. 9x29x+2

  16. 32. 9x2+5x4

  17. 33. 20x2+27x8

  18. 34. 15x219x+6

  19. 35. 2x2+3xy+y2

  20. 36. 3x2+4xy+y2

  21. 37. 6x25xy6y2

  22. 38. 6x27xy5y2

In Exercises 3948, factor the difference of two squares.

  1. 39. x2100

  2. 40. x2144

  3. 41. 36x249

  4. 42. 64x281

  5. 43. 9x225y2

  6. 44. 36x249y2

  7. 45. x416

  8. 46. x41

  9. 47. 16x481

  10. 48. 81x41

In Exercises 4956, factor each perfect square trinomial.

  1. 49. x2+2x+1

  2. 50. x2+4x+4

  3. 51. x214x+49

  4. 52. x210x+25

  5. 53. 4x2+4x+1

  6. 54. 25x2+10x+1

  7. 55. 9x26x+1

  8. 56. 64x216x+1

In Exercises 5764, factor using the formula for the sum or difference of two cubes.

  1. 57. x3+27

  2. 58. x3+64

  3. 59. x364

  4. 60. x327

  5. 61. 8x31

  6. 62. 27x31

  7. 63. 64x3+27

  8. 64. 8x3+125

In Exercises 6592, factor completely, or state that the polynomial is prime.

  1. 65. 3x33x

  2. 66. 5x345x

  3. 67. 4x24x24

  4. 68. 6x218x60

  5. 69. 2x4162

  6. 70. 7x47

  7. 71. x3+2x29x18

  8. 72. x3+3x225x75

  9. 73. 2x22x112

  10. 74. 6x26x12

  11. 75. x34x

  12. 76. 9x39x

  13. 77. x2+64

  14. 78. x2+36

  15. 79. x3+2x24x8

  16. 80. x3+2x2x2

  17. 81. y581y

  18. 82. y516y

  19. 83. 20y445y2

  20. 84. 48y43y2

  21. 85. x212x+3649y2

  22. 86. x210x+2536y2

  23. 87. 9b2 x16y16x+9b2 y

  24. 88. 16a2 x25y25x+16a2 y

  25. 89. x2 y16y+322x2

  26. 90. 12x2 y27y4x2+9

  27. 91. 2x38a2 x+24x2+72x

  28. 92. 2x398a2 x+28x2+98x

In Exercises 93102, factor and simplify each algebraic expression.

  1. 93. x32x12

  2. 94. x34x14

  3. 95. 4x23+8x13

  4. 96. 12x34+6x14

  5. 97. (x+3)12(x+3)32

  6. 98. (x2+4)32+(x2+4)72

  7. 99. (x+5)12(x+5)32

  8. 100. (x2+3)23+(x2+3)53

  9. 101. (4x1)1213(4x1)32

  10. 102. 8(4x+3)2+10(5x+1)(4x+3)1

Practice PLUS

In Exercises 103114, factor completely.

  1. 103. 10x2(x+1)7x(x+1)6(x+1)

  2. 104. 12x2(x1)4x(x1)5(x1)

  3. 105. 6x4+35x26

  4. 106. 7x4+34x25

  5. 107. y7+y

  6. 108. (y+1)3+1

  7. 109. x45x2 y2+4y4

  8. 110. x410x2 y2+9y4

  9. 111. (xy)44(xy)2

  10. 112. (x+y)4100(x+y)2

  11. 113. 2x27xy2+3y4

  12. 114. 3x2+5xy2+2y4

Application Exercises

  1. 115. Your electronics store is having an incredible sale. The price on one laptop is reduced by 40%. Then the sale price is reduced by another 40%. If x is the laptop’s original price, the sale price can be modeled by

    (x0.4x)0.4(x0.4x).
    1. Factor out (x0.4x) from each term. Then simplify the resulting expression.

    2. Use the simplified expression from part (a) to answer these questions. With a 40% reduction followed by a 40% reduction, is the laptop selling at 20% of its original price? If not, at what percentage of the original price is it selling?

  2. 116. Your local electronics store is having an end-of-the-year sale. The price on an Ultra 4K HD television had been reduced by 30%. Now the sale price is reduced by another 30%. If x is the television’s original price, the sale price can be modeled by

    (x0.3x)0.3(x0.3x).
    1. Factor out (x0.3x) from each term. Then simplify the resulting expression.

    2. Use the simplified expression from part (a) to answer these questions. With a 30% reduction followed by a 30% reduction, is the television selling at 40% of its original price? If not, at what percentage of the original price is it selling?

In Exercises 117120,

  1. Write an expression for the area of the shaded region.

  2. Write the expression in factored form.

  1. 117.

    A large squared with sides of length 3 x is divided into 4 smaller squares with sides of length 2, and a plus shaped the central shaded region.
  2. 118.

    A large square with sides of length 7 x is divided into 4 smaller squares at each corner with sides of length 3, and a plus shaped central shaded region within the first square but outside the small squares.
  3. 119.

    A rectangle of length x + y and width x. A smaller rectangle of length x + y and width y is drawn within the first rectangle along its center. The region within the first rectangle but outside the second rectangle is shaded.
  4. 120.

    A diagram is constructed with two squares and a rectangle.

In Exercises 121122, find the formula for the volume of the region outside the smaller rectangular solid and inside the larger rectangular solid. Then express the volume in factored form.

  1. 121.

    A rectangular solid of width b and height b is positioned along the length, at the center, of a larger rectangular solid of length 4 a, width a, and height a.
  2. 122.

    A rectangular solid of width b and height b is positioned along the length, at the center, of a larger rectangular solid of length 3 a, width a, and height a.

Explaining the Concepts

  1. 123. Using an example, explain how to factor out the greatest common factor of a polynomial.

  2. 124. Suppose that a polynomial contains four terms. Explain how to use factoring by grouping to factor the polynomial.

  3. 125. Explain how to factor 3x2+10x+8.

  4. 126. Explain how to factor the difference of two squares. Provide an example with your explanation.

  5. 127. What is a perfect square trinomial and how is it factored?

  6. 128. Explain how to factor x3+1.

  7. 129. What does it mean to factor completely?

Critical Thinking Exercises

Make Sense? In Exercises 130133, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 130. Although 20x3 appears in both 20x3+8x2 and 20x3+10x, I’ll need to factor 20x3 in different ways to obtain each polynomial’s factorization.

  2. 131. You grouped the polynomial’s terms using different groupings than I did, yet we both obtained the same factorization.

  3. 132. I factored 4x2100 completely and obtained (2x+10) (2x10).

  4. 133. First factoring out the greatest common factor makes it easier for me to determine how to factor the remaining factor, assuming that it is not prime.

In Exercises 134137, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 134. x416 is factored completely as (x2+4)(x24).

  2. 135. The trinomial x24x4 is a prime polynomial.

  3. 136. x2+36=(x+6)2

  4. 137. x364=(x+4)(x2+4x16)

In Exercises 138141, factor completely.

  1. 138. x2n+6xn+8

  2. 139. x24x+5

  3. 140. x4y42x3 y+2xy3

  4. 141. (x5)12(x+5)12(x+5)12(x5)32

In Exercises 142143, find all integers b so that the trinomial can be factored.

  1. 142. x2+bx+15

  2. 143. x2+4x+b

Preview Exercises

Exercises 144146 will help you prepare for the material covered in the next section.

  1. 144. Factor the numerator and the denominator. Then simplify by dividing out the common factor in the numerator and the denominator.

    x2+6x+5x225

In Exercises 145146, perform the indicated operation. Where possible, reduce the answer to its lowest terms.

  1. 145. 54815

  2. 146. 12+23

Chapter P: Mid-Chapter Check Point

Chapter P Mid-Chapter Check Point

What You Know: We defined the real numbers [{x|x is rational}{x|x is irrational}] and graphed them as points on a number line. We reviewed the basic rules of algebra, using these properties to simplify algebraic expressions. We expanded our knowledge of exponents to include exponents other than natural numbers:

b0=1;bn=1bn; 1bn=bn;b1n=bn;bmn=(bn)m=bmn;bmn=1bmn.

We used properties of exponents to simplify exponential expressions and properties of radicals to simplify radical expressions. Finally, we performed operations with polynomials. We used a number of fast methods for finding products of polynomials, including the FOIL method for multiplying binomials, a special-product formula for the product of the sum and difference of two terms [(A+B)(AB)=A2B2], and special-product formulas for squaring binomials [(A+B)2=A2+2AB+B2;(AB)2=A22AB+B2]. We reversed the direction of these formulas and reviewed how to factor polynomials. We used a general strategy, summarized in the box here, for factoring a wide variety of polynomials.

In Exercises 127, simplify the given expression or perform the indicated operation (and simplify, if possible), whichever is appropriate.

  1. 1. (3x+5)(4x7)

  2. 2. (3x+5)(4x7)

  3. 3. 6+96

  4. 4. 31227

  5. 5. 7x+3[9(2x6)]

  6. 6. (8x3)2

  7. 7. (x13y12)6

  8. 8. (27)03225

  9. 9. (2x5)(x23x+1)

  10. 10. (2x5)(x23x+1)

  11. 11. x3+x3x3x3

  12. 12. (9a10b)(2a+b)

  13. 13. {a, c, d, e}{c, d, f, h}

  14. 14. {a, c, d, e}{c, d, f, h}

  15. 15. (3x2 y3xy+4y2)(2x2 y33xy+5y2)

  16. 16. 24x2 y132x5 y2

  17. 17. (13 x5 y4)(18x2 y1)

  18. 18. x412(Assume x>0.)

  19. 19. [4y(3x+2)][4y+(3x+2)]

  20. 20. (x2y1)2

  21. 21. 24×1032×106 (Express the answer in scientific notation.)

  22. 22. 32323

  23. 23. (x3+2)(x32)

  24. 24. (x2+2)2

  25. 25. 506

  26. 26. 1173

  27. 27. 113

In Exercises 2834, factor completely, or state that the polynomial is prime.

  1. 28. 7x222x+3

  2. 29. x22x+4

  3. 30. x3+5x2+3x+15

  4. 31. 3x24xy7y2

  5. 32. 64yy4

  6. 33. 50x3+20x2+2x

  7. 34. x26x+949y2

In Exercises 3536, factor and simplify each algebraic expression.

  1. 35. x322x12+x12

  2. 36. (x2+1)1210(x2+1)12

  3. 37. List all the rational numbers in this set:

    {11,37, 0, 0.45, 23,25 }.

In Exercises 3839, rewrite each expression without absolute value bars.

  1. 38. |213|

  2. 39. x2|x|ifx<0

  3. 40. If the population of the United States is approximately 3.3×108 and each person produces about 5 pounds of garbage per day, express the total number of pounds of garbage produced in the United States in one day in scientific notation.

  4. 41. A human brain contains 3×1010 neurons and a gorilla brain contains 7.5×109 neurons. How many times as many neurons are in the brain of a human as in the brain of a gorilla?

  5. 42. College students are graduating with the highest debt burden in history, but that debt may finally be leveling off. The bar graph shows the average student loan debt in the United States for seven selected graduating classes from 2001 through 2019.

    A vertical bar graph titled, average student loan debt in the United States.

    Average student loan debt, D, can be described by the mathematical model

    D=28n2+1352n+15,057

    where n is the number of years after 2000.

    1. Does the mathematical model underestimate or overestimate average student loan debt for the graduating class of 2016? By how much?

    2. If the trend shown by the graph continues, use the formula to project average student loan debt for the graduating class of 2025.

Section P.6: Rational Expressions

Section P.6 Rational Expressions

Learning Objectives

What you’ll learn

  1. 1 Specify numbers that must be excluded from the domain of a rational expression.

  2. 2 Simplify rational expressions.

  3. 3 Multiply rational expressions.

  4. 4 Divide rational expressions.

  5. 5 Add and subtract rational expressions.

  6. 6 Simplify complex rational expressions.

  7. 7 Simplify fractional expressions that occur in calculus.

  8. 8 Rationalize numerators.

We can all do our part to help keep our beaches and riverbanks free of litter and other debris, but what happens when an industrial accident causes toxic chemicals to be discharged into our rivers or an oil spill covers our beaches? The costs of cleaning up after these environmental disasters can be enormous. In this section, we will model such costs using quotients of polynomials.

For example, the algebraic expression

250x100x

describes the cost, in millions of dollars, to remove x percent of the pollutants that are discharged into a river. Removing a modest percentage of pollutants, say 40%, is far less costly than removing a substantially greater percentage, such as 95%. We see this by evaluating the algebraic expression for x=40 and x=95.

Evaluating 250x100xforx=40:  x=95:Cost is 250(40)10040167.Cost is 250(95)10095=4750.

The cost increases from approximately $167 million to a possibly prohibitive $4750 million, or $4.75 billion. Costs spiral upward as the percentage of removed pollutants increases.

While cleanup costs vary based on the size of the disaster, consider this: As of April 2020, ten years after the explosion that caused the Deep Horizon oil spill in the Gulf of Mexico, BP (formerly The British Petroleum Company) had already spent nearly $70 billion on cleanup and settling claims.

Many algebraic expressions that describe costs of environmental projects are examples of rational expressions. First we will define rational expressions. Then we will review how to perform operations with such expressions.

Objective 2: Simplify rational expressions

Simplifying Rational Expressions

  1. Objective 2Simplify rational expressions.

Watch Video

A rational expression is simplified if its numerator and denominator have no common factors other than 1 or 1. The following procedure can be used to simplify rational expressions:

Simplifying Rational Expressions

  1. Factor the numerator and the denominator completely.

  2. Divide both the numerator and the denominator by any common factors.

Example 2 Simplifying Rational Expressions

Simplify:

  1. x3+x2x+1

  2. x2+6x+5x225.

Solution

  1.  

    x3+x2x+1=x2(x+1)x+1Factor the numerator. Because the denominator isx+1, x1.=x2(x+1)1x+11Divide out the common factor,x+1.=x2x1Denominators of 1 need not be written becausea1=a.
  2.  

    x2+6x+5x225=(x+5)(x+1)(x+5)(x5)Factor the numerator and denominator.Because the denominator is(x+5)(x5), x5 and x5.=(x+5)1(x+1)(x+5)1(x5)Divide out the common factor, x+5.=x+1x5,  x5,  x5

Check Point 2

  • Simplify:

    1. x3+3x2x+3

    2. x21x2+2x+1.

Objective 5: Add and subtract rational expressions

Adding and Subtracting Rational Expressions with the Same Denominator

  1. Objective 5Add and subtract rational expressions.

Watch Video

We add or subtract rational expressions with the same denominator by (1) adding or subtracting the numerators, (2) placing this result over the common denominator, and (3) simplifying, if possible.

Example 5 Subtracting Rational Expressions with the Same Denominator

Subtract: 5x+1x294x2x29.

Solution

The process to subtract the rational expressions.

Check Point 5

  • Subtract: xx+13x+2x+1.

Adding and Subtracting Rational Expressions with Different Denominators

We can gain insight into adding rational expressions with different denominators by looking closely at what we do when adding fractions with different denominators. For example, suppose that we want to add 12 and 23. We must first write the fractions with the same denominator. We look for the smallest number that contains both 2 and 3 as factors. This number, 6, is then used as the least common denominator, or LCD.

The least common denominator, or LCD, of several rational expressions is a polynomial consisting of the product of all prime factors in the denominators, with each factor raised to the greatest power of its occurrence in any denominator.

Finding the Least Common Denominator

  1. Factor each denominator completely.

  2. List the factors of the first denominator.

  3. Add to the list in step 2 any factors of the second denominator that do not appear in the list.

  4. Form the product of each different factor from the list in step 3. This product is the least common denominator.

Example 6 Finding the Least Common Denominator

Find the LCD of

x+22x3and4x+3.

Solution

  1. Step 1 FACTOR EACH DENOMINATOR COMPLETELY.

    2 x minus 3 = 1 left parenthesis 2 x minus 3 right parenthesis. x + 3 = 1 left parenthesis x + 3 right parenthesis.
  2. Step 2 LIST THE FACTORS OF THE FIRST DENOMINATOR.

    1, 2x3
  3. Step 3 ADD ANY UNLISTED FACTORS FROM THE SECOND DENOMINATOR. One factor of the second denominator is already in our list. That factor is 1. However, the other factor, x+3, is not listed in step 2. We add x+3 to the list. We have

    1, 2x3, x+3.
  4. Step 4 THE LEAST COMMON DENOMINATOR IS THE PRODUCT OF ALL FACTORS IN THE FINAL LIST. Thus

    1(2x3)(x+3)

    or (2x3)(x+3) is the least common denominator of x+22x3and4x+3.

Check Point 6

  • Find the LCD of

    3x+1 and 5x1.

In Example 6 and Check Point 6, the denominators do not have any common factors other than 1 and 1. Thus, the LCD is simply the product of the denominators. However, in Example 7, the denominators both have a factor of x+3.

Example 7 Finding the Least Common Denominator

Find the LCD of

75x2+15xand9x2+6x+9.

Solution

  1. Step 1 FACTOR EACH DENOMINATOR COMPLETELY.

    5x2+15x=5x(x+3)x2+6x+9=(x+3)2or(x+3)(x+3)
    Start fraction 7 over 5 x squared + 15 x end fraction is labeled, factor are 5, x, and x + 3. Start fraction 9 over x squared + 6 x + 9 end fraction is labeled, factors are x + 3 and x + 3.
  2. Step 2 LIST THE FACTORS OF THE FIRST DENOMINATOR.

    5, x, x+3

  3. Step 3 ADD ANY UNLISTED FACTORS FROM THE SECOND DENOMINATOR. One factor of x2+6x+9 is already in our list. That factor is x+3. However, the other factor of x+3 is not listed in step 2. We add a second factor of x+3 to the list. We have

    5, x, x+3, x+3.
  4. Step 4 THE LEAST COMMON DENOMINATOR IS THE PRODUCT OF ALL FACTORS IN THE FINAL LIST. Thus,

    5x(x+3)(x+3)   or   5x(x+3)2

is the least common denominator.

In Example 7, if you were to multiply the denominators as given, you would get a common denominator, but not the LCD. While any common denominator can be used to add or subtract rational expressions, using the LCD greatly reduces the amount of work involved.

Check Point 7

  • Find the least common denominator of

    3x26x+9and7x29.

Finding the least common denominator for two (or more) rational expressions is the first step needed to add or subtract the expressions.

Adding and Subtracting Rational Expressions That Have Different Denominators

  1. Find the LCD of the rational expressions.

  2. Rewrite each rational expression as an equivalent expression whose denominator is the LCD. To do so, multiply the numerator and the denominator of each rational expression by any factor(s) needed to convert the denominator into the LCD.

  3. Add or subtract numerators, placing the resulting expression over the LCD.

  4. If possible, simplify the resulting rational expression.

Example 8 Adding and Subtracting Rational Expressions with Different Denominators

Add or subtract:

  1. 13+2x

  2. 56x49x2.

Solution

Note that each of these denominators has only one term.

  1. The second rational expression in the sum, 2x, is undefined for x=0, so x0.

  1. Step 1 FIND THE LEAST COMMON DENOMINATOR. The denominators are 3 and x, which have no common factors. Their product, 3x, is the LCD.

  2. Step 2 WRITE EQUIVALENT EXPRESSIONS WITH THE LCD AS DENOMINATORS. We must rewrite each rational expression with a denominator of 3x. We do so by multiplying the numerator and denominator of 13 by x and multiplying the numerator and denominator of 2x by 3.

    The process to write equivalent expressions with the L C D as denominators.

    Because xx=1 and 33=1, we are not changing the value of either rational expression. Now we are ready to perform the addition.

    13+2x  This is the given problem.= x3x+63xReplace each rational expression with its equivalent formcontaining the LCD.
  3. Step 3 ADD NUMERATORS, PLACING THIS SUM OVER THE LCD.

    = x+63x,x0Add numerators and write the restriction on the variable.
  4. Step 4 Simplify, if necessary. Because the numerator is prime, no further simplification is possible.

  1. Both rational expressions in the subtraction 56x49x2 are undefined for x=0, so x0.

  1. Step 1 Find the least common denominator. The denominators are 6x and 9x2. Begin by listing the factors of 6x:

    2, 3, x.

    The factors of 9x2 are 3, 3, x, and x. The list of factors of 6x only contains one factor of 3 and one factor of x, so we add one more 3 and one more x to the list:

    2, 3, 3, x, x.

    The LCD is the product of the factors in this list. The LCD is 18x2.

  2. Step 2 Write equivalent expressions with the LCD as denominators. We must rewrite each rational expression with a denominator of 18x2. We do so by multiplying the numerator and denominator of 56x by 3x and multiplying the numerator and denominator of 49x2 by 2.

    The process to write equivalent expressions with the L C D as denominators.

    Because 3x3x=1 and 22=1, we are not changing the value of either rational expression. Now we are ready to perform the subtraction.

    56x49x2This is the given problem.= 15x18x2818x2Replace each rational expression with its equivalentfrom containing the LCD.
  3. Step 3 SUBTRACT NUMERATORS, PLACING THIS SUM OVER THE LCD.

    = 15x818x2,x0Subtract numerators and write the restriction on the variable.
  4. Step 4 Simplify, if necessary. Because the numerator is prime, no further simplification is possible.

Check Point 8

  • Add or subtract:

    1. x+158x

    2. 710x3+54x2.

Example 9 Subtracting Rational Expressions with Different Denominators

Subtract: x+22x34x+3.

Solution

  1. Step 1 FIND THE LEAST COMMON DENOMINATOR. In Example 6, we found that the LCD for these rational expressions is (2x3)(x+3).

  2. Step 2 WRITE EQUIVALENT EXPRESSIONS WITH THE LCD AS DENOMINATORS. We must rewrite each rational expression with a denominator of (2x3)(x+3). We do so by multiplying both the numerator and the denominator of each rational expression by any factor needed to convert the expression’s denominator into the LCD.

    The process to write equivalent expressions with the L C D as denominators.

    Because x+3x+3=1 and 2x32x3=1, we are not changing the value of either rational expression, only its appearance.

    Now we are ready to perform the indicated subtraction.

    x+22x34x+3This is the given problem.=(x+2)(x+3)(2x3)(x+3)4(2x3)(x+3)(2x3)Multiply each numerator and denominatorby the extra factor required to form(2x3)(x+3), the LCD.
  3. Step 3 SUBTRACT NUMERATORS, PUTTING THIS DIFFERENCE OVER THE LCD.

    =(x+2)(x+3)4(2x3)(2x3)(x+3)=x2+5x+6(8x12)(2x3)(x+3)Multiply in the numerator using FOIL andthe distributive property.=x2+5x+68x+12(2x3)(x+3)Remove parentheses and then change thesign of each term in parentheses.=x23x+18(2x3)(x+3),x32,x3Combine like terms in the numerator.
  4. Step 4 IF NECESSARY, SIMPLIFY. Because the numerator is prime, no further simplification is possible.

Check Point 9

  • Add: xx3+x1x+3.

Example 10 Adding Rational Expressions with Different Denominators

Add: x+3x2+x2+2x21.

Solution

  1. Step 1 FIND THE LEAST COMMON DENOMINATOR. Start by factoring the denominators.

    x2+x2=(x+2)(x1)x21=(x+1)(x1)

    The factors of the first denominator are x+2 and x1. The only factor from the second denominator that is not listed is x+1. Thus, the least common denominator is

    (x+2)(x1)(x+1).
  2. Step 2 WRITE EQUIVALENT EXPRESSIONS WITH THE LCD AS DENOMINATORS. We must rewrite each rational expression with a denominator of (x+2)(x1)(x+1). We do so by multiplying both the numerator and the denominator of each rational expression by any factor(s) needed to convert the expression’s denominator into the LCD.

    The process to write equivalent expressions with the L C D as denominators.

    Because x+1x+1=1 and x+2x+2=1, we are not changing the value of either rational expression, only its appearance.

    Now we are ready to perform the indicated addition.

    x+3x2+x2+2x21This is the given problem.=x+3(x+2)(x1)+2(x+1)(x1)Factor the denominators.The LCD is(x+2)(x1)(x+1).=(x+3)(x+1)(x+2)(x1)(x+1)+2(x+2)(x+2)(x1)(x+1)Rewrite as equivalentexpressions with the LCD.
  3. Step 3 ADD NUMERATORS, PUTTING THIS SUM OVER THE LCD.

    =(x+3)(x+1)+2(x+2)(x+2)(x1)(x+1)=x2+4x+3+2x+4(x+2)(x1)(x+1)Perform the multiplicationsin the numerator.=x2+6x+7(x+2)(x1)(x+1), x2, x1, x1Combine like terms in thenumerator:  4x+2x=6xand  3+4=7.
  4. Step 4 IF NECESSARY, SIMPLIFY. Because the numerator is prime, no further simplification is possible.

Check Point 10

  • Subtract: xx210x+25x42x10.

Objective 5: Add and subtract rational expressions

Adding and Subtracting Rational Expressions with the Same Denominator

  1. Objective 5Add and subtract rational expressions.

Watch Video

We add or subtract rational expressions with the same denominator by (1) adding or subtracting the numerators, (2) placing this result over the common denominator, and (3) simplifying, if possible.

Example 5 Subtracting Rational Expressions with the Same Denominator

Subtract: 5x+1x294x2x29.

Solution

The process to subtract the rational expressions.

Check Point 5

  • Subtract: xx+13x+2x+1.

Adding and Subtracting Rational Expressions with Different Denominators

We can gain insight into adding rational expressions with different denominators by looking closely at what we do when adding fractions with different denominators. For example, suppose that we want to add 12 and 23. We must first write the fractions with the same denominator. We look for the smallest number that contains both 2 and 3 as factors. This number, 6, is then used as the least common denominator, or LCD.

The least common denominator, or LCD, of several rational expressions is a polynomial consisting of the product of all prime factors in the denominators, with each factor raised to the greatest power of its occurrence in any denominator.

Finding the Least Common Denominator

  1. Factor each denominator completely.

  2. List the factors of the first denominator.

  3. Add to the list in step 2 any factors of the second denominator that do not appear in the list.

  4. Form the product of each different factor from the list in step 3. This product is the least common denominator.

Example 6 Finding the Least Common Denominator

Find the LCD of

x+22x3and4x+3.

Solution

  1. Step 1 FACTOR EACH DENOMINATOR COMPLETELY.

    2 x minus 3 = 1 left parenthesis 2 x minus 3 right parenthesis. x + 3 = 1 left parenthesis x + 3 right parenthesis.
  2. Step 2 LIST THE FACTORS OF THE FIRST DENOMINATOR.

    1, 2x3
  3. Step 3 ADD ANY UNLISTED FACTORS FROM THE SECOND DENOMINATOR. One factor of the second denominator is already in our list. That factor is 1. However, the other factor, x+3, is not listed in step 2. We add x+3 to the list. We have

    1, 2x3, x+3.
  4. Step 4 THE LEAST COMMON DENOMINATOR IS THE PRODUCT OF ALL FACTORS IN THE FINAL LIST. Thus

    1(2x3)(x+3)

    or (2x3)(x+3) is the least common denominator of x+22x3and4x+3.

Check Point 6

  • Find the LCD of

    3x+1 and 5x1.

In Example 6 and Check Point 6, the denominators do not have any common factors other than 1 and 1. Thus, the LCD is simply the product of the denominators. However, in Example 7, the denominators both have a factor of x+3.

Example 7 Finding the Least Common Denominator

Find the LCD of

75x2+15xand9x2+6x+9.

Solution

  1. Step 1 FACTOR EACH DENOMINATOR COMPLETELY.

    5x2+15x=5x(x+3)x2+6x+9=(x+3)2or(x+3)(x+3)
    Start fraction 7 over 5 x squared + 15 x end fraction is labeled, factor are 5, x, and x + 3. Start fraction 9 over x squared + 6 x + 9 end fraction is labeled, factors are x + 3 and x + 3.
  2. Step 2 LIST THE FACTORS OF THE FIRST DENOMINATOR.

    5, x, x+3

  3. Step 3 ADD ANY UNLISTED FACTORS FROM THE SECOND DENOMINATOR. One factor of x2+6x+9 is already in our list. That factor is x+3. However, the other factor of x+3 is not listed in step 2. We add a second factor of x+3 to the list. We have

    5, x, x+3, x+3.
  4. Step 4 THE LEAST COMMON DENOMINATOR IS THE PRODUCT OF ALL FACTORS IN THE FINAL LIST. Thus,

    5x(x+3)(x+3)   or   5x(x+3)2

is the least common denominator.

In Example 7, if you were to multiply the denominators as given, you would get a common denominator, but not the LCD. While any common denominator can be used to add or subtract rational expressions, using the LCD greatly reduces the amount of work involved.

Check Point 7

  • Find the least common denominator of

    3x26x+9and7x29.

Finding the least common denominator for two (or more) rational expressions is the first step needed to add or subtract the expressions.

Adding and Subtracting Rational Expressions That Have Different Denominators

  1. Find the LCD of the rational expressions.

  2. Rewrite each rational expression as an equivalent expression whose denominator is the LCD. To do so, multiply the numerator and the denominator of each rational expression by any factor(s) needed to convert the denominator into the LCD.

  3. Add or subtract numerators, placing the resulting expression over the LCD.

  4. If possible, simplify the resulting rational expression.

Example 8 Adding and Subtracting Rational Expressions with Different Denominators

Add or subtract:

  1. 13+2x

  2. 56x49x2.

Solution

Note that each of these denominators has only one term.

  1. The second rational expression in the sum, 2x, is undefined for x=0, so x0.

  1. Step 1 FIND THE LEAST COMMON DENOMINATOR. The denominators are 3 and x, which have no common factors. Their product, 3x, is the LCD.

  2. Step 2 WRITE EQUIVALENT EXPRESSIONS WITH THE LCD AS DENOMINATORS. We must rewrite each rational expression with a denominator of 3x. We do so by multiplying the numerator and denominator of 13 by x and multiplying the numerator and denominator of 2x by 3.

    The process to write equivalent expressions with the L C D as denominators.

    Because xx=1 and 33=1, we are not changing the value of either rational expression. Now we are ready to perform the addition.

    13+2x  This is the given problem.= x3x+63xReplace each rational expression with its equivalent formcontaining the LCD.
  3. Step 3 ADD NUMERATORS, PLACING THIS SUM OVER THE LCD.

    = x+63x,x0Add numerators and write the restriction on the variable.
  4. Step 4 Simplify, if necessary. Because the numerator is prime, no further simplification is possible.

  1. Both rational expressions in the subtraction 56x49x2 are undefined for x=0, so x0.

  1. Step 1 Find the least common denominator. The denominators are 6x and 9x2. Begin by listing the factors of 6x:

    2, 3, x.

    The factors of 9x2 are 3, 3, x, and x. The list of factors of 6x only contains one factor of 3 and one factor of x, so we add one more 3 and one more x to the list:

    2, 3, 3, x, x.

    The LCD is the product of the factors in this list. The LCD is 18x2.

  2. Step 2 Write equivalent expressions with the LCD as denominators. We must rewrite each rational expression with a denominator of 18x2. We do so by multiplying the numerator and denominator of 56x by 3x and multiplying the numerator and denominator of 49x2 by 2.

    The process to write equivalent expressions with the L C D as denominators.

    Because 3x3x=1 and 22=1, we are not changing the value of either rational expression. Now we are ready to perform the subtraction.

    56x49x2This is the given problem.= 15x18x2818x2Replace each rational expression with its equivalentfrom containing the LCD.
  3. Step 3 SUBTRACT NUMERATORS, PLACING THIS SUM OVER THE LCD.

    = 15x818x2,x0Subtract numerators and write the restriction on the variable.
  4. Step 4 Simplify, if necessary. Because the numerator is prime, no further simplification is possible.

Check Point 8

  • Add or subtract:

    1. x+158x

    2. 710x3+54x2.

Example 9 Subtracting Rational Expressions with Different Denominators

Subtract: x+22x34x+3.

Solution

  1. Step 1 FIND THE LEAST COMMON DENOMINATOR. In Example 6, we found that the LCD for these rational expressions is (2x3)(x+3).

  2. Step 2 WRITE EQUIVALENT EXPRESSIONS WITH THE LCD AS DENOMINATORS. We must rewrite each rational expression with a denominator of (2x3)(x+3). We do so by multiplying both the numerator and the denominator of each rational expression by any factor needed to convert the expression’s denominator into the LCD.

    The process to write equivalent expressions with the L C D as denominators.

    Because x+3x+3=1 and 2x32x3=1, we are not changing the value of either rational expression, only its appearance.

    Now we are ready to perform the indicated subtraction.

    x+22x34x+3This is the given problem.=(x+2)(x+3)(2x3)(x+3)4(2x3)(x+3)(2x3)Multiply each numerator and denominatorby the extra factor required to form(2x3)(x+3), the LCD.
  3. Step 3 SUBTRACT NUMERATORS, PUTTING THIS DIFFERENCE OVER THE LCD.

    =(x+2)(x+3)4(2x3)(2x3)(x+3)=x2+5x+6(8x12)(2x3)(x+3)Multiply in the numerator using FOIL andthe distributive property.=x2+5x+68x+12(2x3)(x+3)Remove parentheses and then change thesign of each term in parentheses.=x23x+18(2x3)(x+3),x32,x3Combine like terms in the numerator.
  4. Step 4 IF NECESSARY, SIMPLIFY. Because the numerator is prime, no further simplification is possible.

Check Point 9

  • Add: xx3+x1x+3.

Example 10 Adding Rational Expressions with Different Denominators

Add: x+3x2+x2+2x21.

Solution

  1. Step 1 FIND THE LEAST COMMON DENOMINATOR. Start by factoring the denominators.

    x2+x2=(x+2)(x1)x21=(x+1)(x1)

    The factors of the first denominator are x+2 and x1. The only factor from the second denominator that is not listed is x+1. Thus, the least common denominator is

    (x+2)(x1)(x+1).
  2. Step 2 WRITE EQUIVALENT EXPRESSIONS WITH THE LCD AS DENOMINATORS. We must rewrite each rational expression with a denominator of (x+2)(x1)(x+1). We do so by multiplying both the numerator and the denominator of each rational expression by any factor(s) needed to convert the expression’s denominator into the LCD.

    The process to write equivalent expressions with the L C D as denominators.

    Because x+1x+1=1 and x+2x+2=1, we are not changing the value of either rational expression, only its appearance.

    Now we are ready to perform the indicated addition.

    x+3x2+x2+2x21This is the given problem.=x+3(x+2)(x1)+2(x+1)(x1)Factor the denominators.The LCD is(x+2)(x1)(x+1).=(x+3)(x+1)(x+2)(x1)(x+1)+2(x+2)(x+2)(x1)(x+1)Rewrite as equivalentexpressions with the LCD.
  3. Step 3 ADD NUMERATORS, PUTTING THIS SUM OVER THE LCD.

    =(x+3)(x+1)+2(x+2)(x+2)(x1)(x+1)=x2+4x+3+2x+4(x+2)(x1)(x+1)Perform the multiplicationsin the numerator.=x2+6x+7(x+2)(x1)(x+1), x2, x1, x1Combine like terms in thenumerator:  4x+2x=6xand  3+4=7.
  4. Step 4 IF NECESSARY, SIMPLIFY. Because the numerator is prime, no further simplification is possible.

Check Point 10

  • Subtract: xx210x+25x42x10.

Objective 5: Add and subtract rational expressions

Adding and Subtracting Rational Expressions with the Same Denominator

  1. Objective 5Add and subtract rational expressions.

Watch Video

We add or subtract rational expressions with the same denominator by (1) adding or subtracting the numerators, (2) placing this result over the common denominator, and (3) simplifying, if possible.

Example 5 Subtracting Rational Expressions with the Same Denominator

Subtract: 5x+1x294x2x29.

Solution

The process to subtract the rational expressions.

Check Point 5

  • Subtract: xx+13x+2x+1.

Adding and Subtracting Rational Expressions with Different Denominators

We can gain insight into adding rational expressions with different denominators by looking closely at what we do when adding fractions with different denominators. For example, suppose that we want to add 12 and 23. We must first write the fractions with the same denominator. We look for the smallest number that contains both 2 and 3 as factors. This number, 6, is then used as the least common denominator, or LCD.

The least common denominator, or LCD, of several rational expressions is a polynomial consisting of the product of all prime factors in the denominators, with each factor raised to the greatest power of its occurrence in any denominator.

Finding the Least Common Denominator

  1. Factor each denominator completely.

  2. List the factors of the first denominator.

  3. Add to the list in step 2 any factors of the second denominator that do not appear in the list.

  4. Form the product of each different factor from the list in step 3. This product is the least common denominator.

Example 6 Finding the Least Common Denominator

Find the LCD of

x+22x3and4x+3.

Solution

  1. Step 1 FACTOR EACH DENOMINATOR COMPLETELY.

    2 x minus 3 = 1 left parenthesis 2 x minus 3 right parenthesis. x + 3 = 1 left parenthesis x + 3 right parenthesis.
  2. Step 2 LIST THE FACTORS OF THE FIRST DENOMINATOR.

    1, 2x3
  3. Step 3 ADD ANY UNLISTED FACTORS FROM THE SECOND DENOMINATOR. One factor of the second denominator is already in our list. That factor is 1. However, the other factor, x+3, is not listed in step 2. We add x+3 to the list. We have

    1, 2x3, x+3.
  4. Step 4 THE LEAST COMMON DENOMINATOR IS THE PRODUCT OF ALL FACTORS IN THE FINAL LIST. Thus

    1(2x3)(x+3)

    or (2x3)(x+3) is the least common denominator of x+22x3and4x+3.

Check Point 6

  • Find the LCD of

    3x+1 and 5x1.

In Example 6 and Check Point 6, the denominators do not have any common factors other than 1 and 1. Thus, the LCD is simply the product of the denominators. However, in Example 7, the denominators both have a factor of x+3.

Example 7 Finding the Least Common Denominator

Find the LCD of

75x2+15xand9x2+6x+9.

Solution

  1. Step 1 FACTOR EACH DENOMINATOR COMPLETELY.

    5x2+15x=5x(x+3)x2+6x+9=(x+3)2or(x+3)(x+3)
    Start fraction 7 over 5 x squared + 15 x end fraction is labeled, factor are 5, x, and x + 3. Start fraction 9 over x squared + 6 x + 9 end fraction is labeled, factors are x + 3 and x + 3.
  2. Step 2 LIST THE FACTORS OF THE FIRST DENOMINATOR.

    5, x, x+3

  3. Step 3 ADD ANY UNLISTED FACTORS FROM THE SECOND DENOMINATOR. One factor of x2+6x+9 is already in our list. That factor is x+3. However, the other factor of x+3 is not listed in step 2. We add a second factor of x+3 to the list. We have

    5, x, x+3, x+3.
  4. Step 4 THE LEAST COMMON DENOMINATOR IS THE PRODUCT OF ALL FACTORS IN THE FINAL LIST. Thus,

    5x(x+3)(x+3)   or   5x(x+3)2

is the least common denominator.

In Example 7, if you were to multiply the denominators as given, you would get a common denominator, but not the LCD. While any common denominator can be used to add or subtract rational expressions, using the LCD greatly reduces the amount of work involved.

Check Point 7

  • Find the least common denominator of

    3x26x+9and7x29.

Finding the least common denominator for two (or more) rational expressions is the first step needed to add or subtract the expressions.

Adding and Subtracting Rational Expressions That Have Different Denominators

  1. Find the LCD of the rational expressions.

  2. Rewrite each rational expression as an equivalent expression whose denominator is the LCD. To do so, multiply the numerator and the denominator of each rational expression by any factor(s) needed to convert the denominator into the LCD.

  3. Add or subtract numerators, placing the resulting expression over the LCD.

  4. If possible, simplify the resulting rational expression.

Example 8 Adding and Subtracting Rational Expressions with Different Denominators

Add or subtract:

  1. 13+2x

  2. 56x49x2.

Solution

Note that each of these denominators has only one term.

  1. The second rational expression in the sum, 2x, is undefined for x=0, so x0.

  1. Step 1 FIND THE LEAST COMMON DENOMINATOR. The denominators are 3 and x, which have no common factors. Their product, 3x, is the LCD.

  2. Step 2 WRITE EQUIVALENT EXPRESSIONS WITH THE LCD AS DENOMINATORS. We must rewrite each rational expression with a denominator of 3x. We do so by multiplying the numerator and denominator of 13 by x and multiplying the numerator and denominator of 2x by 3.

    The process to write equivalent expressions with the L C D as denominators.

    Because xx=1 and 33=1, we are not changing the value of either rational expression. Now we are ready to perform the addition.

    13+2x  This is the given problem.= x3x+63xReplace each rational expression with its equivalent formcontaining the LCD.
  3. Step 3 ADD NUMERATORS, PLACING THIS SUM OVER THE LCD.

    = x+63x,x0Add numerators and write the restriction on the variable.
  4. Step 4 Simplify, if necessary. Because the numerator is prime, no further simplification is possible.

  1. Both rational expressions in the subtraction 56x49x2 are undefined for x=0, so x0.

  1. Step 1 Find the least common denominator. The denominators are 6x and 9x2. Begin by listing the factors of 6x:

    2, 3, x.

    The factors of 9x2 are 3, 3, x, and x. The list of factors of 6x only contains one factor of 3 and one factor of x, so we add one more 3 and one more x to the list:

    2, 3, 3, x, x.

    The LCD is the product of the factors in this list. The LCD is 18x2.

  2. Step 2 Write equivalent expressions with the LCD as denominators. We must rewrite each rational expression with a denominator of 18x2. We do so by multiplying the numerator and denominator of 56x by 3x and multiplying the numerator and denominator of 49x2 by 2.

    The process to write equivalent expressions with the L C D as denominators.

    Because 3x3x=1 and 22=1, we are not changing the value of either rational expression. Now we are ready to perform the subtraction.

    56x49x2This is the given problem.= 15x18x2818x2Replace each rational expression with its equivalentfrom containing the LCD.
  3. Step 3 SUBTRACT NUMERATORS, PLACING THIS SUM OVER THE LCD.

    = 15x818x2,x0Subtract numerators and write the restriction on the variable.
  4. Step 4 Simplify, if necessary. Because the numerator is prime, no further simplification is possible.

Check Point 8

  • Add or subtract:

    1. x+158x

    2. 710x3+54x2.

Example 9 Subtracting Rational Expressions with Different Denominators

Subtract: x+22x34x+3.

Solution

  1. Step 1 FIND THE LEAST COMMON DENOMINATOR. In Example 6, we found that the LCD for these rational expressions is (2x3)(x+3).

  2. Step 2 WRITE EQUIVALENT EXPRESSIONS WITH THE LCD AS DENOMINATORS. We must rewrite each rational expression with a denominator of (2x3)(x+3). We do so by multiplying both the numerator and the denominator of each rational expression by any factor needed to convert the expression’s denominator into the LCD.

    The process to write equivalent expressions with the L C D as denominators.

    Because x+3x+3=1 and 2x32x3=1, we are not changing the value of either rational expression, only its appearance.

    Now we are ready to perform the indicated subtraction.

    x+22x34x+3This is the given problem.=(x+2)(x+3)(2x3)(x+3)4(2x3)(x+3)(2x3)Multiply each numerator and denominatorby the extra factor required to form(2x3)(x+3), the LCD.
  3. Step 3 SUBTRACT NUMERATORS, PUTTING THIS DIFFERENCE OVER THE LCD.

    =(x+2)(x+3)4(2x3)(2x3)(x+3)=x2+5x+6(8x12)(2x3)(x+3)Multiply in the numerator using FOIL andthe distributive property.=x2+5x+68x+12(2x3)(x+3)Remove parentheses and then change thesign of each term in parentheses.=x23x+18(2x3)(x+3),x32,x3Combine like terms in the numerator.
  4. Step 4 IF NECESSARY, SIMPLIFY. Because the numerator is prime, no further simplification is possible.

Check Point 9

  • Add: xx3+x1x+3.

Example 10 Adding Rational Expressions with Different Denominators

Add: x+3x2+x2+2x21.

Solution

  1. Step 1 FIND THE LEAST COMMON DENOMINATOR. Start by factoring the denominators.

    x2+x2=(x+2)(x1)x21=(x+1)(x1)

    The factors of the first denominator are x+2 and x1. The only factor from the second denominator that is not listed is x+1. Thus, the least common denominator is

    (x+2)(x1)(x+1).
  2. Step 2 WRITE EQUIVALENT EXPRESSIONS WITH THE LCD AS DENOMINATORS. We must rewrite each rational expression with a denominator of (x+2)(x1)(x+1). We do so by multiplying both the numerator and the denominator of each rational expression by any factor(s) needed to convert the expression’s denominator into the LCD.

    The process to write equivalent expressions with the L C D as denominators.

    Because x+1x+1=1 and x+2x+2=1, we are not changing the value of either rational expression, only its appearance.

    Now we are ready to perform the indicated addition.

    x+3x2+x2+2x21This is the given problem.=x+3(x+2)(x1)+2(x+1)(x1)Factor the denominators.The LCD is(x+2)(x1)(x+1).=(x+3)(x+1)(x+2)(x1)(x+1)+2(x+2)(x+2)(x1)(x+1)Rewrite as equivalentexpressions with the LCD.
  3. Step 3 ADD NUMERATORS, PUTTING THIS SUM OVER THE LCD.

    =(x+3)(x+1)+2(x+2)(x+2)(x1)(x+1)=x2+4x+3+2x+4(x+2)(x1)(x+1)Perform the multiplicationsin the numerator.=x2+6x+7(x+2)(x1)(x+1), x2, x1, x1Combine like terms in thenumerator:  4x+2x=6xand  3+4=7.
  4. Step 4 IF NECESSARY, SIMPLIFY. Because the numerator is prime, no further simplification is possible.

Check Point 10

  • Subtract: xx210x+25x42x10.

Objective 5: Add and subtract rational expressions

Adding and Subtracting Rational Expressions with the Same Denominator

  1. Objective 5Add and subtract rational expressions.

Watch Video

We add or subtract rational expressions with the same denominator by (1) adding or subtracting the numerators, (2) placing this result over the common denominator, and (3) simplifying, if possible.

Example 5 Subtracting Rational Expressions with the Same Denominator

Subtract: 5x+1x294x2x29.

Solution

The process to subtract the rational expressions.

Check Point 5

  • Subtract: xx+13x+2x+1.

Adding and Subtracting Rational Expressions with Different Denominators

We can gain insight into adding rational expressions with different denominators by looking closely at what we do when adding fractions with different denominators. For example, suppose that we want to add 12 and 23. We must first write the fractions with the same denominator. We look for the smallest number that contains both 2 and 3 as factors. This number, 6, is then used as the least common denominator, or LCD.

The least common denominator, or LCD, of several rational expressions is a polynomial consisting of the product of all prime factors in the denominators, with each factor raised to the greatest power of its occurrence in any denominator.

Finding the Least Common Denominator

  1. Factor each denominator completely.

  2. List the factors of the first denominator.

  3. Add to the list in step 2 any factors of the second denominator that do not appear in the list.

  4. Form the product of each different factor from the list in step 3. This product is the least common denominator.

Example 6 Finding the Least Common Denominator

Find the LCD of

x+22x3and4x+3.

Solution

  1. Step 1 FACTOR EACH DENOMINATOR COMPLETELY.

    2 x minus 3 = 1 left parenthesis 2 x minus 3 right parenthesis. x + 3 = 1 left parenthesis x + 3 right parenthesis.
  2. Step 2 LIST THE FACTORS OF THE FIRST DENOMINATOR.

    1, 2x3
  3. Step 3 ADD ANY UNLISTED FACTORS FROM THE SECOND DENOMINATOR. One factor of the second denominator is already in our list. That factor is 1. However, the other factor, x+3, is not listed in step 2. We add x+3 to the list. We have

    1, 2x3, x+3.
  4. Step 4 THE LEAST COMMON DENOMINATOR IS THE PRODUCT OF ALL FACTORS IN THE FINAL LIST. Thus

    1(2x3)(x+3)

    or (2x3)(x+3) is the least common denominator of x+22x3and4x+3.

Check Point 6

  • Find the LCD of

    3x+1 and 5x1.

In Example 6 and Check Point 6, the denominators do not have any common factors other than 1 and 1. Thus, the LCD is simply the product of the denominators. However, in Example 7, the denominators both have a factor of x+3.

Example 7 Finding the Least Common Denominator

Find the LCD of

75x2+15xand9x2+6x+9.

Solution

  1. Step 1 FACTOR EACH DENOMINATOR COMPLETELY.

    5x2+15x=5x(x+3)x2+6x+9=(x+3)2or(x+3)(x+3)
    Start fraction 7 over 5 x squared + 15 x end fraction is labeled, factor are 5, x, and x + 3. Start fraction 9 over x squared + 6 x + 9 end fraction is labeled, factors are x + 3 and x + 3.
  2. Step 2 LIST THE FACTORS OF THE FIRST DENOMINATOR.

    5, x, x+3

  3. Step 3 ADD ANY UNLISTED FACTORS FROM THE SECOND DENOMINATOR. One factor of x2+6x+9 is already in our list. That factor is x+3. However, the other factor of x+3 is not listed in step 2. We add a second factor of x+3 to the list. We have

    5, x, x+3, x+3.
  4. Step 4 THE LEAST COMMON DENOMINATOR IS THE PRODUCT OF ALL FACTORS IN THE FINAL LIST. Thus,

    5x(x+3)(x+3)   or   5x(x+3)2

is the least common denominator.

In Example 7, if you were to multiply the denominators as given, you would get a common denominator, but not the LCD. While any common denominator can be used to add or subtract rational expressions, using the LCD greatly reduces the amount of work involved.

Check Point 7

  • Find the least common denominator of

    3x26x+9and7x29.

Finding the least common denominator for two (or more) rational expressions is the first step needed to add or subtract the expressions.

Adding and Subtracting Rational Expressions That Have Different Denominators

  1. Find the LCD of the rational expressions.

  2. Rewrite each rational expression as an equivalent expression whose denominator is the LCD. To do so, multiply the numerator and the denominator of each rational expression by any factor(s) needed to convert the denominator into the LCD.

  3. Add or subtract numerators, placing the resulting expression over the LCD.

  4. If possible, simplify the resulting rational expression.

Example 8 Adding and Subtracting Rational Expressions with Different Denominators

Add or subtract:

  1. 13+2x

  2. 56x49x2.

Solution

Note that each of these denominators has only one term.

  1. The second rational expression in the sum, 2x, is undefined for x=0, so x0.

  1. Step 1 FIND THE LEAST COMMON DENOMINATOR. The denominators are 3 and x, which have no common factors. Their product, 3x, is the LCD.

  2. Step 2 WRITE EQUIVALENT EXPRESSIONS WITH THE LCD AS DENOMINATORS. We must rewrite each rational expression with a denominator of 3x. We do so by multiplying the numerator and denominator of 13 by x and multiplying the numerator and denominator of 2x by 3.

    The process to write equivalent expressions with the L C D as denominators.

    Because xx=1 and 33=1, we are not changing the value of either rational expression. Now we are ready to perform the addition.

    13+2x  This is the given problem.= x3x+63xReplace each rational expression with its equivalent formcontaining the LCD.
  3. Step 3 ADD NUMERATORS, PLACING THIS SUM OVER THE LCD.

    = x+63x,x0Add numerators and write the restriction on the variable.
  4. Step 4 Simplify, if necessary. Because the numerator is prime, no further simplification is possible.

  1. Both rational expressions in the subtraction 56x49x2 are undefined for x=0, so x0.

  1. Step 1 Find the least common denominator. The denominators are 6x and 9x2. Begin by listing the factors of 6x:

    2, 3, x.

    The factors of 9x2 are 3, 3, x, and x. The list of factors of 6x only contains one factor of 3 and one factor of x, so we add one more 3 and one more x to the list:

    2, 3, 3, x, x.

    The LCD is the product of the factors in this list. The LCD is 18x2.

  2. Step 2 Write equivalent expressions with the LCD as denominators. We must rewrite each rational expression with a denominator of 18x2. We do so by multiplying the numerator and denominator of 56x by 3x and multiplying the numerator and denominator of 49x2 by 2.

    The process to write equivalent expressions with the L C D as denominators.

    Because 3x3x=1 and 22=1, we are not changing the value of either rational expression. Now we are ready to perform the subtraction.

    56x49x2This is the given problem.= 15x18x2818x2Replace each rational expression with its equivalentfrom containing the LCD.
  3. Step 3 SUBTRACT NUMERATORS, PLACING THIS SUM OVER THE LCD.

    = 15x818x2,x0Subtract numerators and write the restriction on the variable.
  4. Step 4 Simplify, if necessary. Because the numerator is prime, no further simplification is possible.

Check Point 8

  • Add or subtract:

    1. x+158x

    2. 710x3+54x2.

Example 9 Subtracting Rational Expressions with Different Denominators

Subtract: x+22x34x+3.

Solution

  1. Step 1 FIND THE LEAST COMMON DENOMINATOR. In Example 6, we found that the LCD for these rational expressions is (2x3)(x+3).

  2. Step 2 WRITE EQUIVALENT EXPRESSIONS WITH THE LCD AS DENOMINATORS. We must rewrite each rational expression with a denominator of (2x3)(x+3). We do so by multiplying both the numerator and the denominator of each rational expression by any factor needed to convert the expression’s denominator into the LCD.

    The process to write equivalent expressions with the L C D as denominators.

    Because x+3x+3=1 and 2x32x3=1, we are not changing the value of either rational expression, only its appearance.

    Now we are ready to perform the indicated subtraction.

    x+22x34x+3This is the given problem.=(x+2)(x+3)(2x3)(x+3)4(2x3)(x+3)(2x3)Multiply each numerator and denominatorby the extra factor required to form(2x3)(x+3), the LCD.
  3. Step 3 SUBTRACT NUMERATORS, PUTTING THIS DIFFERENCE OVER THE LCD.

    =(x+2)(x+3)4(2x3)(2x3)(x+3)=x2+5x+6(8x12)(2x3)(x+3)Multiply in the numerator using FOIL andthe distributive property.=x2+5x+68x+12(2x3)(x+3)Remove parentheses and then change thesign of each term in parentheses.=x23x+18(2x3)(x+3),x32,x3Combine like terms in the numerator.
  4. Step 4 IF NECESSARY, SIMPLIFY. Because the numerator is prime, no further simplification is possible.

Check Point 9

  • Add: xx3+x1x+3.

Example 10 Adding Rational Expressions with Different Denominators

Add: x+3x2+x2+2x21.

Solution

  1. Step 1 FIND THE LEAST COMMON DENOMINATOR. Start by factoring the denominators.

    x2+x2=(x+2)(x1)x21=(x+1)(x1)

    The factors of the first denominator are x+2 and x1. The only factor from the second denominator that is not listed is x+1. Thus, the least common denominator is

    (x+2)(x1)(x+1).
  2. Step 2 WRITE EQUIVALENT EXPRESSIONS WITH THE LCD AS DENOMINATORS. We must rewrite each rational expression with a denominator of (x+2)(x1)(x+1). We do so by multiplying both the numerator and the denominator of each rational expression by any factor(s) needed to convert the expression’s denominator into the LCD.

    The process to write equivalent expressions with the L C D as denominators.

    Because x+1x+1=1 and x+2x+2=1, we are not changing the value of either rational expression, only its appearance.

    Now we are ready to perform the indicated addition.

    x+3x2+x2+2x21This is the given problem.=x+3(x+2)(x1)+2(x+1)(x1)Factor the denominators.The LCD is(x+2)(x1)(x+1).=(x+3)(x+1)(x+2)(x1)(x+1)+2(x+2)(x+2)(x1)(x+1)Rewrite as equivalentexpressions with the LCD.
  3. Step 3 ADD NUMERATORS, PUTTING THIS SUM OVER THE LCD.

    =(x+3)(x+1)+2(x+2)(x+2)(x1)(x+1)=x2+4x+3+2x+4(x+2)(x1)(x+1)Perform the multiplicationsin the numerator.=x2+6x+7(x+2)(x1)(x+1), x2, x1, x1Combine like terms in thenumerator:  4x+2x=6xand  3+4=7.
  4. Step 4 IF NECESSARY, SIMPLIFY. Because the numerator is prime, no further simplification is possible.

Check Point 10

  • Subtract: xx210x+25x42x10.

Objective 5: Add and subtract rational expressions

Adding and Subtracting Rational Expressions with the Same Denominator

  1. Objective 5Add and subtract rational expressions.

Watch Video

We add or subtract rational expressions with the same denominator by (1) adding or subtracting the numerators, (2) placing this result over the common denominator, and (3) simplifying, if possible.

Example 5 Subtracting Rational Expressions with the Same Denominator

Subtract: 5x+1x294x2x29.

Solution

The process to subtract the rational expressions.

Check Point 5

  • Subtract: xx+13x+2x+1.

Adding and Subtracting Rational Expressions with Different Denominators

We can gain insight into adding rational expressions with different denominators by looking closely at what we do when adding fractions with different denominators. For example, suppose that we want to add 12 and 23. We must first write the fractions with the same denominator. We look for the smallest number that contains both 2 and 3 as factors. This number, 6, is then used as the least common denominator, or LCD.

The least common denominator, or LCD, of several rational expressions is a polynomial consisting of the product of all prime factors in the denominators, with each factor raised to the greatest power of its occurrence in any denominator.

Finding the Least Common Denominator

  1. Factor each denominator completely.

  2. List the factors of the first denominator.

  3. Add to the list in step 2 any factors of the second denominator that do not appear in the list.

  4. Form the product of each different factor from the list in step 3. This product is the least common denominator.

Example 6 Finding the Least Common Denominator

Find the LCD of

x+22x3and4x+3.

Solution

  1. Step 1 FACTOR EACH DENOMINATOR COMPLETELY.

    2 x minus 3 = 1 left parenthesis 2 x minus 3 right parenthesis. x + 3 = 1 left parenthesis x + 3 right parenthesis.
  2. Step 2 LIST THE FACTORS OF THE FIRST DENOMINATOR.

    1, 2x3
  3. Step 3 ADD ANY UNLISTED FACTORS FROM THE SECOND DENOMINATOR. One factor of the second denominator is already in our list. That factor is 1. However, the other factor, x+3, is not listed in step 2. We add x+3 to the list. We have

    1, 2x3, x+3.
  4. Step 4 THE LEAST COMMON DENOMINATOR IS THE PRODUCT OF ALL FACTORS IN THE FINAL LIST. Thus

    1(2x3)(x+3)

    or (2x3)(x+3) is the least common denominator of x+22x3and4x+3.

Check Point 6

  • Find the LCD of

    3x+1 and 5x1.

In Example 6 and Check Point 6, the denominators do not have any common factors other than 1 and 1. Thus, the LCD is simply the product of the denominators. However, in Example 7, the denominators both have a factor of x+3.

Example 7 Finding the Least Common Denominator

Find the LCD of

75x2+15xand9x2+6x+9.

Solution

  1. Step 1 FACTOR EACH DENOMINATOR COMPLETELY.

    5x2+15x=5x(x+3)x2+6x+9=(x+3)2or(x+3)(x+3)
    Start fraction 7 over 5 x squared + 15 x end fraction is labeled, factor are 5, x, and x + 3. Start fraction 9 over x squared + 6 x + 9 end fraction is labeled, factors are x + 3 and x + 3.
  2. Step 2 LIST THE FACTORS OF THE FIRST DENOMINATOR.

    5, x, x+3

  3. Step 3 ADD ANY UNLISTED FACTORS FROM THE SECOND DENOMINATOR. One factor of x2+6x+9 is already in our list. That factor is x+3. However, the other factor of x+3 is not listed in step 2. We add a second factor of x+3 to the list. We have

    5, x, x+3, x+3.
  4. Step 4 THE LEAST COMMON DENOMINATOR IS THE PRODUCT OF ALL FACTORS IN THE FINAL LIST. Thus,

    5x(x+3)(x+3)   or   5x(x+3)2

is the least common denominator.

In Example 7, if you were to multiply the denominators as given, you would get a common denominator, but not the LCD. While any common denominator can be used to add or subtract rational expressions, using the LCD greatly reduces the amount of work involved.

Check Point 7

  • Find the least common denominator of

    3x26x+9and7x29.

Finding the least common denominator for two (or more) rational expressions is the first step needed to add or subtract the expressions.

Adding and Subtracting Rational Expressions That Have Different Denominators

  1. Find the LCD of the rational expressions.

  2. Rewrite each rational expression as an equivalent expression whose denominator is the LCD. To do so, multiply the numerator and the denominator of each rational expression by any factor(s) needed to convert the denominator into the LCD.

  3. Add or subtract numerators, placing the resulting expression over the LCD.

  4. If possible, simplify the resulting rational expression.

Example 8 Adding and Subtracting Rational Expressions with Different Denominators

Add or subtract:

  1. 13+2x

  2. 56x49x2.

Solution

Note that each of these denominators has only one term.

  1. The second rational expression in the sum, 2x, is undefined for x=0, so x0.

  1. Step 1 FIND THE LEAST COMMON DENOMINATOR. The denominators are 3 and x, which have no common factors. Their product, 3x, is the LCD.

  2. Step 2 WRITE EQUIVALENT EXPRESSIONS WITH THE LCD AS DENOMINATORS. We must rewrite each rational expression with a denominator of 3x. We do so by multiplying the numerator and denominator of 13 by x and multiplying the numerator and denominator of 2x by 3.

    The process to write equivalent expressions with the L C D as denominators.

    Because xx=1 and 33=1, we are not changing the value of either rational expression. Now we are ready to perform the addition.

    13+2x  This is the given problem.= x3x+63xReplace each rational expression with its equivalent formcontaining the LCD.
  3. Step 3 ADD NUMERATORS, PLACING THIS SUM OVER THE LCD.

    = x+63x,x0Add numerators and write the restriction on the variable.
  4. Step 4 Simplify, if necessary. Because the numerator is prime, no further simplification is possible.

  1. Both rational expressions in the subtraction 56x49x2 are undefined for x=0, so x0.

  1. Step 1 Find the least common denominator. The denominators are 6x and 9x2. Begin by listing the factors of 6x:

    2, 3, x.

    The factors of 9x2 are 3, 3, x, and x. The list of factors of 6x only contains one factor of 3 and one factor of x, so we add one more 3 and one more x to the list:

    2, 3, 3, x, x.

    The LCD is the product of the factors in this list. The LCD is 18x2.

  2. Step 2 Write equivalent expressions with the LCD as denominators. We must rewrite each rational expression with a denominator of 18x2. We do so by multiplying the numerator and denominator of 56x by 3x and multiplying the numerator and denominator of 49x2 by 2.

    The process to write equivalent expressions with the L C D as denominators.

    Because 3x3x=1 and 22=1, we are not changing the value of either rational expression. Now we are ready to perform the subtraction.

    56x49x2This is the given problem.= 15x18x2818x2Replace each rational expression with its equivalentfrom containing the LCD.
  3. Step 3 SUBTRACT NUMERATORS, PLACING THIS SUM OVER THE LCD.

    = 15x818x2,x0Subtract numerators and write the restriction on the variable.
  4. Step 4 Simplify, if necessary. Because the numerator is prime, no further simplification is possible.

Check Point 8

  • Add or subtract:

    1. x+158x

    2. 710x3+54x2.

Example 9 Subtracting Rational Expressions with Different Denominators

Subtract: x+22x34x+3.

Solution

  1. Step 1 FIND THE LEAST COMMON DENOMINATOR. In Example 6, we found that the LCD for these rational expressions is (2x3)(x+3).

  2. Step 2 WRITE EQUIVALENT EXPRESSIONS WITH THE LCD AS DENOMINATORS. We must rewrite each rational expression with a denominator of (2x3)(x+3). We do so by multiplying both the numerator and the denominator of each rational expression by any factor needed to convert the expression’s denominator into the LCD.

    The process to write equivalent expressions with the L C D as denominators.

    Because x+3x+3=1 and 2x32x3=1, we are not changing the value of either rational expression, only its appearance.

    Now we are ready to perform the indicated subtraction.

    x+22x34x+3This is the given problem.=(x+2)(x+3)(2x3)(x+3)4(2x3)(x+3)(2x3)Multiply each numerator and denominatorby the extra factor required to form(2x3)(x+3), the LCD.
  3. Step 3 SUBTRACT NUMERATORS, PUTTING THIS DIFFERENCE OVER THE LCD.

    =(x+2)(x+3)4(2x3)(2x3)(x+3)=x2+5x+6(8x12)(2x3)(x+3)Multiply in the numerator using FOIL andthe distributive property.=x2+5x+68x+12(2x3)(x+3)Remove parentheses and then change thesign of each term in parentheses.=x23x+18(2x3)(x+3),x32,x3Combine like terms in the numerator.
  4. Step 4 IF NECESSARY, SIMPLIFY. Because the numerator is prime, no further simplification is possible.

Check Point 9

  • Add: xx3+x1x+3.

Example 10 Adding Rational Expressions with Different Denominators

Add: x+3x2+x2+2x21.

Solution

  1. Step 1 FIND THE LEAST COMMON DENOMINATOR. Start by factoring the denominators.

    x2+x2=(x+2)(x1)x21=(x+1)(x1)

    The factors of the first denominator are x+2 and x1. The only factor from the second denominator that is not listed is x+1. Thus, the least common denominator is

    (x+2)(x1)(x+1).
  2. Step 2 WRITE EQUIVALENT EXPRESSIONS WITH THE LCD AS DENOMINATORS. We must rewrite each rational expression with a denominator of (x+2)(x1)(x+1). We do so by multiplying both the numerator and the denominator of each rational expression by any factor(s) needed to convert the expression’s denominator into the LCD.

    The process to write equivalent expressions with the L C D as denominators.

    Because x+1x+1=1 and x+2x+2=1, we are not changing the value of either rational expression, only its appearance.

    Now we are ready to perform the indicated addition.

    x+3x2+x2+2x21This is the given problem.=x+3(x+2)(x1)+2(x+1)(x1)Factor the denominators.The LCD is(x+2)(x1)(x+1).=(x+3)(x+1)(x+2)(x1)(x+1)+2(x+2)(x+2)(x1)(x+1)Rewrite as equivalentexpressions with the LCD.
  3. Step 3 ADD NUMERATORS, PUTTING THIS SUM OVER THE LCD.

    =(x+3)(x+1)+2(x+2)(x+2)(x1)(x+1)=x2+4x+3+2x+4(x+2)(x1)(x+1)Perform the multiplicationsin the numerator.=x2+6x+7(x+2)(x1)(x+1), x2, x1, x1Combine like terms in thenumerator:  4x+2x=6xand  3+4=7.
  4. Step 4 IF NECESSARY, SIMPLIFY. Because the numerator is prime, no further simplification is possible.

Check Point 10

  • Subtract: xx210x+25x42x10.

Objective 5: Add and subtract rational expressions

Adding and Subtracting Rational Expressions with the Same Denominator

  1. Objective 5Add and subtract rational expressions.

Watch Video

We add or subtract rational expressions with the same denominator by (1) adding or subtracting the numerators, (2) placing this result over the common denominator, and (3) simplifying, if possible.

Example 5 Subtracting Rational Expressions with the Same Denominator

Subtract: 5x+1x294x2x29.

Solution

The process to subtract the rational expressions.

Check Point 5

  • Subtract: xx+13x+2x+1.

Adding and Subtracting Rational Expressions with Different Denominators

We can gain insight into adding rational expressions with different denominators by looking closely at what we do when adding fractions with different denominators. For example, suppose that we want to add 12 and 23. We must first write the fractions with the same denominator. We look for the smallest number that contains both 2 and 3 as factors. This number, 6, is then used as the least common denominator, or LCD.

The least common denominator, or LCD, of several rational expressions is a polynomial consisting of the product of all prime factors in the denominators, with each factor raised to the greatest power of its occurrence in any denominator.

Finding the Least Common Denominator

  1. Factor each denominator completely.

  2. List the factors of the first denominator.

  3. Add to the list in step 2 any factors of the second denominator that do not appear in the list.

  4. Form the product of each different factor from the list in step 3. This product is the least common denominator.

Example 6 Finding the Least Common Denominator

Find the LCD of

x+22x3and4x+3.

Solution

  1. Step 1 FACTOR EACH DENOMINATOR COMPLETELY.

    2 x minus 3 = 1 left parenthesis 2 x minus 3 right parenthesis. x + 3 = 1 left parenthesis x + 3 right parenthesis.
  2. Step 2 LIST THE FACTORS OF THE FIRST DENOMINATOR.

    1, 2x3
  3. Step 3 ADD ANY UNLISTED FACTORS FROM THE SECOND DENOMINATOR. One factor of the second denominator is already in our list. That factor is 1. However, the other factor, x+3, is not listed in step 2. We add x+3 to the list. We have

    1, 2x3, x+3.
  4. Step 4 THE LEAST COMMON DENOMINATOR IS THE PRODUCT OF ALL FACTORS IN THE FINAL LIST. Thus

    1(2x3)(x+3)

    or (2x3)(x+3) is the least common denominator of x+22x3and4x+3.

Check Point 6

  • Find the LCD of

    3x+1 and 5x1.

In Example 6 and Check Point 6, the denominators do not have any common factors other than 1 and 1. Thus, the LCD is simply the product of the denominators. However, in Example 7, the denominators both have a factor of x+3.

Example 7 Finding the Least Common Denominator

Find the LCD of

75x2+15xand9x2+6x+9.

Solution

  1. Step 1 FACTOR EACH DENOMINATOR COMPLETELY.

    5x2+15x=5x(x+3)x2+6x+9=(x+3)2or(x+3)(x+3)
    Start fraction 7 over 5 x squared + 15 x end fraction is labeled, factor are 5, x, and x + 3. Start fraction 9 over x squared + 6 x + 9 end fraction is labeled, factors are x + 3 and x + 3.
  2. Step 2 LIST THE FACTORS OF THE FIRST DENOMINATOR.

    5, x, x+3

  3. Step 3 ADD ANY UNLISTED FACTORS FROM THE SECOND DENOMINATOR. One factor of x2+6x+9 is already in our list. That factor is x+3. However, the other factor of x+3 is not listed in step 2. We add a second factor of x+3 to the list. We have

    5, x, x+3, x+3.
  4. Step 4 THE LEAST COMMON DENOMINATOR IS THE PRODUCT OF ALL FACTORS IN THE FINAL LIST. Thus,

    5x(x+3)(x+3)   or   5x(x+3)2

is the least common denominator.

In Example 7, if you were to multiply the denominators as given, you would get a common denominator, but not the LCD. While any common denominator can be used to add or subtract rational expressions, using the LCD greatly reduces the amount of work involved.

Check Point 7

  • Find the least common denominator of

    3x26x+9and7x29.

Finding the least common denominator for two (or more) rational expressions is the first step needed to add or subtract the expressions.

Adding and Subtracting Rational Expressions That Have Different Denominators

  1. Find the LCD of the rational expressions.

  2. Rewrite each rational expression as an equivalent expression whose denominator is the LCD. To do so, multiply the numerator and the denominator of each rational expression by any factor(s) needed to convert the denominator into the LCD.

  3. Add or subtract numerators, placing the resulting expression over the LCD.

  4. If possible, simplify the resulting rational expression.

Example 8 Adding and Subtracting Rational Expressions with Different Denominators

Add or subtract:

  1. 13+2x

  2. 56x49x2.

Solution

Note that each of these denominators has only one term.

  1. The second rational expression in the sum, 2x, is undefined for x=0, so x0.

  1. Step 1 FIND THE LEAST COMMON DENOMINATOR. The denominators are 3 and x, which have no common factors. Their product, 3x, is the LCD.

  2. Step 2 WRITE EQUIVALENT EXPRESSIONS WITH THE LCD AS DENOMINATORS. We must rewrite each rational expression with a denominator of 3x. We do so by multiplying the numerator and denominator of 13 by x and multiplying the numerator and denominator of 2x by 3.

    The process to write equivalent expressions with the L C D as denominators.

    Because xx=1 and 33=1, we are not changing the value of either rational expression. Now we are ready to perform the addition.

    13+2x  This is the given problem.= x3x+63xReplace each rational expression with its equivalent formcontaining the LCD.
  3. Step 3 ADD NUMERATORS, PLACING THIS SUM OVER THE LCD.

    = x+63x,x0Add numerators and write the restriction on the variable.
  4. Step 4 Simplify, if necessary. Because the numerator is prime, no further simplification is possible.

  1. Both rational expressions in the subtraction 56x49x2 are undefined for x=0, so x0.

  1. Step 1 Find the least common denominator. The denominators are 6x and 9x2. Begin by listing the factors of 6x:

    2, 3, x.

    The factors of 9x2 are 3, 3, x, and x. The list of factors of 6x only contains one factor of 3 and one factor of x, so we add one more 3 and one more x to the list:

    2, 3, 3, x, x.

    The LCD is the product of the factors in this list. The LCD is 18x2.

  2. Step 2 Write equivalent expressions with the LCD as denominators. We must rewrite each rational expression with a denominator of 18x2. We do so by multiplying the numerator and denominator of 56x by 3x and multiplying the numerator and denominator of 49x2 by 2.

    The process to write equivalent expressions with the L C D as denominators.

    Because 3x3x=1 and 22=1, we are not changing the value of either rational expression. Now we are ready to perform the subtraction.

    56x49x2This is the given problem.= 15x18x2818x2Replace each rational expression with its equivalentfrom containing the LCD.
  3. Step 3 SUBTRACT NUMERATORS, PLACING THIS SUM OVER THE LCD.

    = 15x818x2,x0Subtract numerators and write the restriction on the variable.
  4. Step 4 Simplify, if necessary. Because the numerator is prime, no further simplification is possible.

Check Point 8

  • Add or subtract:

    1. x+158x

    2. 710x3+54x2.

Example 9 Subtracting Rational Expressions with Different Denominators

Subtract: x+22x34x+3.

Solution

  1. Step 1 FIND THE LEAST COMMON DENOMINATOR. In Example 6, we found that the LCD for these rational expressions is (2x3)(x+3).

  2. Step 2 WRITE EQUIVALENT EXPRESSIONS WITH THE LCD AS DENOMINATORS. We must rewrite each rational expression with a denominator of (2x3)(x+3). We do so by multiplying both the numerator and the denominator of each rational expression by any factor needed to convert the expression’s denominator into the LCD.

    The process to write equivalent expressions with the L C D as denominators.

    Because x+3x+3=1 and 2x32x3=1, we are not changing the value of either rational expression, only its appearance.

    Now we are ready to perform the indicated subtraction.

    x+22x34x+3This is the given problem.=(x+2)(x+3)(2x3)(x+3)4(2x3)(x+3)(2x3)Multiply each numerator and denominatorby the extra factor required to form(2x3)(x+3), the LCD.
  3. Step 3 SUBTRACT NUMERATORS, PUTTING THIS DIFFERENCE OVER THE LCD.

    =(x+2)(x+3)4(2x3)(2x3)(x+3)=x2+5x+6(8x12)(2x3)(x+3)Multiply in the numerator using FOIL andthe distributive property.=x2+5x+68x+12(2x3)(x+3)Remove parentheses and then change thesign of each term in parentheses.=x23x+18(2x3)(x+3),x32,x3Combine like terms in the numerator.
  4. Step 4 IF NECESSARY, SIMPLIFY. Because the numerator is prime, no further simplification is possible.

Check Point 9

  • Add: xx3+x1x+3.

Example 10 Adding Rational Expressions with Different Denominators

Add: x+3x2+x2+2x21.

Solution

  1. Step 1 FIND THE LEAST COMMON DENOMINATOR. Start by factoring the denominators.

    x2+x2=(x+2)(x1)x21=(x+1)(x1)

    The factors of the first denominator are x+2 and x1. The only factor from the second denominator that is not listed is x+1. Thus, the least common denominator is

    (x+2)(x1)(x+1).
  2. Step 2 WRITE EQUIVALENT EXPRESSIONS WITH THE LCD AS DENOMINATORS. We must rewrite each rational expression with a denominator of (x+2)(x1)(x+1). We do so by multiplying both the numerator and the denominator of each rational expression by any factor(s) needed to convert the expression’s denominator into the LCD.

    The process to write equivalent expressions with the L C D as denominators.

    Because x+1x+1=1 and x+2x+2=1, we are not changing the value of either rational expression, only its appearance.

    Now we are ready to perform the indicated addition.

    x+3x2+x2+2x21This is the given problem.=x+3(x+2)(x1)+2(x+1)(x1)Factor the denominators.The LCD is(x+2)(x1)(x+1).=(x+3)(x+1)(x+2)(x1)(x+1)+2(x+2)(x+2)(x1)(x+1)Rewrite as equivalentexpressions with the LCD.
  3. Step 3 ADD NUMERATORS, PUTTING THIS SUM OVER THE LCD.

    =(x+3)(x+1)+2(x+2)(x+2)(x1)(x+1)=x2+4x+3+2x+4(x+2)(x1)(x+1)Perform the multiplicationsin the numerator.=x2+6x+7(x+2)(x1)(x+1), x2, x1, x1Combine like terms in thenumerator:  4x+2x=6xand  3+4=7.
  4. Step 4 IF NECESSARY, SIMPLIFY. Because the numerator is prime, no further simplification is possible.

Check Point 10

  • Subtract: xx210x+25x42x10.

Objective 6: Simplify complex rational expressions

Complex Rational Expressions

  1. Objective 6Simplify complex rational expressions.

Watch Video

Complex rational expressions, also called complex fractions, have numerators or denominators containing one or more rational expressions. Here are two examples of such expressions:

Examples of complex rational expressions.
P.7-152 Full Alternative Text

One method for simplifying a complex rational expression is to combine its numerator into a single expression and combine its denominator into a single expression. Then perform the division by inverting the denominator and multiplying.

Example 11 Simplifying a Complex Rational Expression

Simplify: 1+1x11x.

Solution

  1. Step 1 ADD TO GET A SINGLE RATIONAL EXPRESSION IN THE NUMERATOR.

    The process to simplify a complex rational expression.
  2. Step 2 SUBTRACT TO GET A SINGLE RATIONAL EXPRESSION IN THE DENOMINATOR.

    The process to simplify a complex rational expression.
  3. Step 3 PERFORM THE DIVISION INDICATED BY THE MAIN FRACTION BAR: INVERT AND MULTIPLY. IF POSSIBLE, SIMPLIFY.

    The process to simplify a complex rational expression.

Check Point 11

  • Simplify: 1x321x+34.

A second method for simplifying a complex rational expression is to find the least common denominator of all the rational expressions in its numerator and denominator. Then multiply each term in its numerator and denominator by this least common denominator. Because we are multiplying by a form of 1, we will obtain an equivalent expression that does not contain fractions in its numerator or denominator. Here we use this method to simplify the complex rational expression in Example 11.

1+1x11x=(1+1x)(11x)xxThe least common denominator of all the rationalexpressions is x. Multiply the numerator anddenominator by x. Becausexx=1,we are notchanging the complex fraction (x0).=1x+1xx1x1xxUse the distributive property. Be sure to distribute  x  toevery term.=x+1x1, x0, x1Multiply. The complex rational expression is now simplified.

Example 12 Simplifying a Complex Rational Expression

Simplify: 1x+h1xh.

Solution

We will use the method of multiplying each of the three terms, 1x+h1x, and h, by the least common denominator. The least common denominator is x(x+h).

1x+h1xh=(1x+h1x)x(x+h)hx(x+h)Multiply the numerator and denominator byx(x+h), h0, x0, xh.=1x+hx(x+h)1xx(x+h)hx(x+h)Use the distributive property in the numerator.=x(x+h)hx(x+h)Simplify:1x+hx(x+h)=x and 1xx(x+h)=x+h.=xxhhx(x+h)Subtract in the numerator. Remove parenthesesand change the sign of each term in parentheses.=h1h1x(x+h)Simplify:  xxh=h.=1x(x+h), h0, x0, xhDivide the numerator and denominator by  h.

Check Point 12

  • Simplify: 1x+71x7.

Objective 6: Simplify complex rational expressions

Complex Rational Expressions

  1. Objective 6Simplify complex rational expressions.

Watch Video

Complex rational expressions, also called complex fractions, have numerators or denominators containing one or more rational expressions. Here are two examples of such expressions:

Examples of complex rational expressions.
P.7-152 Full Alternative Text

One method for simplifying a complex rational expression is to combine its numerator into a single expression and combine its denominator into a single expression. Then perform the division by inverting the denominator and multiplying.

Example 11 Simplifying a Complex Rational Expression

Simplify: 1+1x11x.

Solution

  1. Step 1 ADD TO GET A SINGLE RATIONAL EXPRESSION IN THE NUMERATOR.

    The process to simplify a complex rational expression.
  2. Step 2 SUBTRACT TO GET A SINGLE RATIONAL EXPRESSION IN THE DENOMINATOR.

    The process to simplify a complex rational expression.
  3. Step 3 PERFORM THE DIVISION INDICATED BY THE MAIN FRACTION BAR: INVERT AND MULTIPLY. IF POSSIBLE, SIMPLIFY.

    The process to simplify a complex rational expression.

Check Point 11

  • Simplify: 1x321x+34.

A second method for simplifying a complex rational expression is to find the least common denominator of all the rational expressions in its numerator and denominator. Then multiply each term in its numerator and denominator by this least common denominator. Because we are multiplying by a form of 1, we will obtain an equivalent expression that does not contain fractions in its numerator or denominator. Here we use this method to simplify the complex rational expression in Example 11.

1+1x11x=(1+1x)(11x)xxThe least common denominator of all the rationalexpressions is x. Multiply the numerator anddenominator by x. Becausexx=1,we are notchanging the complex fraction (x0).=1x+1xx1x1xxUse the distributive property. Be sure to distribute  x  toevery term.=x+1x1, x0, x1Multiply. The complex rational expression is now simplified.

Example 12 Simplifying a Complex Rational Expression

Simplify: 1x+h1xh.

Solution

We will use the method of multiplying each of the three terms, 1x+h1x, and h, by the least common denominator. The least common denominator is x(x+h).

1x+h1xh=(1x+h1x)x(x+h)hx(x+h)Multiply the numerator and denominator byx(x+h), h0, x0, xh.=1x+hx(x+h)1xx(x+h)hx(x+h)Use the distributive property in the numerator.=x(x+h)hx(x+h)Simplify:1x+hx(x+h)=x and 1xx(x+h)=x+h.=xxhhx(x+h)Subtract in the numerator. Remove parenthesesand change the sign of each term in parentheses.=h1h1x(x+h)Simplify:  xxh=h.=1x(x+h), h0, x0, xhDivide the numerator and denominator by  h.

Check Point 12

  • Simplify: 1x+71x7.

Objective 7: Simplify fractional expressions that occur in calculus

Fractional Expressions in Calculus

  1. Objective 7Simplify fractional expressions that occur in calculus.

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Fractional expressions containing radicals occur frequently in calculus. Because of the radicals, these expressions are not rational expressions. However, they can often be simplified using the procedure for simplifying complex rational expressions.

Example 13 Simplifying a Fractional Expression Containing Radicals

Simplify: 9x2+x29x29x2.

Solution

9x2+x29x29x2The least common denominator is9x2. =9x2+x29x29x29x2 9x2Multiply the numerator and thedenominator by 9x2. =9x2 9x2+x29x29x2(9x2)9x2Use the distributive property in thenumerator. =(9x2)+x2(9x2)32In the denominator:(9x2)1(9x2)12=(9x2)1+12=(9x2)32. =9(9x2)3Because the original expression wasin radical form,write the denominatorin radical form.

Check Point 13

  • Simplify: x+1xx.

Objective 8: Rationalize numerators

  1. Objective 8Rationalize numerators.

Watch Video

Another fractional expression that you will encounter in calculus is

x+hxh.

Can you see that this expression is not defined if h=0? However, in calculus, you will ask the following question:

The question is answered by first rationalizing the numerator. This process involves rewriting the fractional expression as an equivalent expression in which the numerator no longer contains any radicals. To rationalize a numerator, multiply by 1 to eliminate the radicals in the numerator. Multiply the numerator and the denominator by the conjugate of the numerator.

Example 14 Rationalizing a Numerator

Rationalize the numerator:

x+hxh.

Solution

The conjugate of the numerator is x+h+x. If we multiply the numerator and denominator by x+h+x, the simplified numerator will not contain a radical. Therefore, we multiply by 1, choosing x+h+xx+h+x for 1.

x+hxh=x+hxhx+h+xx+h+xMultiplyby1.=(x+h)2 (x)2 hx+h+x(ab)(a+b)=(a)2(b)2=x+hxhx+h+x(x+h)2=x+hand(x)2=x.=hh(x+h+x)Simplify: x+hx=h.=1x+h+x,h0Divide both the numerator anddenominator by h.

What happens to x+hxh as h gets closer and closer to 0? In Example 14, we showed that

x+hxh=1x+h+x.

As h gets closer to 0, the expression on the right gets closer to 1x+0+x=1x+x, or 12x. Thus, the fractional expression x+hxh approaches 12x as h gets closer to 0.

Check Point 14

  • Rationalize the numerator: x+3x3.

P.6: Concept and Vocabulary Check

P.6 Concept and Vocabulary Check

Fill in each blank so that the resulting statement is true.

  1. C1. A rational expression is the quotient of two _________.

  2. C2. The set of real numbers for which a rational expression is defined is the _________ of the expression. We must exclude all numbers from this set that make the denominator of the rational expression ______.

  3. C3. We simplify a rational expression by _________ the numerator and the denominator completely. Then we divide the numerator and the denominator by any ______________.

  4. C4. x5x3= _______

  5. C5. x5÷x3= _______, x0

  6. C6. x23x43= _______

  7. C7. Consider the following subtraction problem:

    x1x2+x6x2x2+4x+3.

    The factors of the first denominator are _____________.

    The factors of the second denominator are ________.

    The LCD is ____________.

  8. C8. An equivalent expression for 3x+2x5 with a denominator of (3x+4)(x5) can be obtained by multiplying the numerator and denominator by ______.

  9. C9. A rational expression whose numerator or denominator or both contain rational expressions is called a/an ________ rational expression or a/an ________ fraction.

  10. C10. 1x+31x3=x(x+3)x(x+3)(1x+31x)3=_(____)3x(x+3)=____3x(x+3)=____________

  11. C11. We can simplify

    x+1xx

    by multiplying the numerator and the denominator by ______.

  12. C12. We can rationalize the numerator of x+7x7 by multiplying the numerator and the denominator by ______.

P.6: Exercise Set

P.6 Exercise Set

Practice Exercises

In Exercises 16, find all numbers that must be excluded from the domain of each rational expression.

  1. 1. 7x3

  2. 2. 13x+9

  3. 3. x+5x225

  4. 4. x+7x249

  5. 5. x1x2+11x+10

  6. 6. x3x2+4x45

In Exercises 714, simplify each rational expression. Find all numbers that must be excluded from the domain of the simplified rational expression.

  1. 7. 3x9x26x+9

  2. 8. 4x8x24x+4

  3. 9. x212x+364x24

  4. 10. x28x+163x12

  5. 11. y2+7y18y23y+2

  6. 12. y24y5y2+5y+4

  7. 13. x2+12x+36x236

  8. 14. x214x+49x249

In Exercises 1532, multiply or divide as indicated.

  1. 15. x23x+92x+62x4

  2. 16. 6x+93x15x54x+6

  3. 17. x29x2x23xx2+x12

  4. 18. x24x24x+42x4x+2

  5. 19. x25x+6x22x3x21x24

  6. 20. x2+5x+6x2+x6x29x2x6

  7. 21. x38x24x+23x

  8. 22. x2+6x+9x3+271x+3

  9. 23. x+13÷3x+37

  10. 24. x+57÷4x+209

  11. 25. x24x÷x+2x2

  12. 26. x24x2÷x+24x8

  13. 27. 4x2+10x3÷6x2+15x29

  14. 28. x2+xx24÷x21x2+5x+6

  15. 29. x2252x2÷x2+10x+25x2+4x5

  16. 30. x24x2+3x10÷x2+5x+6x2+8x+15

  17. 31. x2+x12x2+x30x2+5x+6x22x3÷x+3x2+7x+6

  18. 32. x325x4x22x22x26x+5÷x2+5x7x+7

In Exercises 3368, add or subtract as indicated.

  1. 33. 4x+16x+5+8x+96x+5

  2. 34. 3x+23x+4+3x+63x+4

  3. 35. x22xx2+3x+x2+xx2+3x

  4. 36. x24xx2x6+4x4x2x6

  5. 37. 4x10x2x4x2

  6. 38. 2x+33x63x3x6

  7. 39. x2+3xx2+x12x212x2+x12

  8. 40. x24xx2x6x6x2x6

  9. 41. 5x+3

  10. 42. 74x

  11. 43. x4+5x6

  12. 44. 3x8+x12

  13. 45. 103x83

  14. 46. 7565x

  15. 47. 25xx+14x

  16. 48. 47x+x13x

  17. 49. 56x+x38x2

  18. 50. x+910x3+1115x2

  19. 51. 3x+4+6x+5

  20. 52. 8x2+2x3

  21. 53. 3x+13x

  22. 54. 4x3x+3

  23. 55. 2xx+2+x+2x2

  24. 56. 3xx3x+4x+2

  25. 57. x+5x5+x5x+5

  26. 58. x+3x3+x3x+3

  27. 59. 32x+4+23x+6

  28. 60. 52x+8+73x+12

  29. 61. 4x2+6x+9+4x+3

  30. 62. 35x+2+5x25x24

  31. 63. 3xx2+3x102xx2+x6

  32. 64. xx22x24xx27x+6

  33. 65. x+3x21x+2x1

  34. 66. x+5x24x+1x2

  35. 67. 4x2+x6x2+3x+23xx+1+5x+2

  36. 68. 6x2+17x40x2+x20+3x45xx+5

In Exercises 6982, simplify each complex rational expression.

  1. 69. x31x3

  2. 70. x41x4

  3. 71. 1+1x31x

  4. 72. 8+1x41x

  5. 73. 1x+1yx+y

  6. 74. 11xxy

  7. 75. xxx+3x+2

  8. 76. x3x3x2

  9. 77. 3x24x+27x24

  10. 78. xx2+13x24+1

  11. 79. 1x+11x22x3+1x3

  12. 80. 6x2+2x151x31x+5+1

  13. 81. 1(x+h)21x2h

  14. 82. x+hx+h+1xx+1h

Exercises 83–88 contain fractional expressions that occur frequently in calculus. Simplify each expression.

  1. 83. x13xx

  2. 84. x14xx

  3. 85. x2x2+2x2+2x2

  4. 86. 5x2+x25x25x2

  5. 87. 1x+h1xh

  6. 88. 1x+31x3

In Exercises 89–92, rationalize the numerator.

  1. 89. x+5x5

  2. 90. x+7x7

  3. 91. x+yx2y2

  4. 92. xyx2y2

Practice PLUS

In Exercises 93100, perform the indicated operations. Simplify the result, if possible.

  1. 93. (2x+3x+1x2+4x52x2+x3)2x+2

  2. 94. 1x22x8÷(1x41x+2)

  3. 95. (26x+1)(1+3x2)

  4. 96. (43x+2)(1+5x1)

  5. 97. y1(y+5)15

  6. 98. y1(y+2)12

  7. 99. (1a3b3ac+adbcbd1)cda2+ab+b2

  8. 100. aba2+ab+b2+(acadbc+bdacad+bcbd÷a3b3a3+b3)

Application Exercises

  1. 101. The rational expression

    130x100x

    describes the cost, in millions of dollars, to inoculate x percent of the population against a particular strain of flu.

    1. Evaluate the expression for x=40, x=80, and x=90. Describe the meaning of each evaluation in terms of percentage inoculated and cost.

    2. For what value of x is the expression undefined?

    3. What happens to the cost as x approaches 100%? How can you interpret this observation?

  2. 102. The average rate on a round-trip commute having a one-way distance d is given by the complex rational expression

    2ddr1+dr2,

    in which r1 and r2 are the average rates on the outgoing and return trips, respectively. Simplify the expression. Then find your average rate if you drive to campus averaging 40 miles per hour and return home on the same route averaging 30 miles per hour. Explain why the answer is not 35 miles per hour.

  3. 103. The bar graph shows the estimated number of calories per day needed to maintain energy balance for various gender and age groups for moderately active lifestyles. (Moderately active means a lifestyle that includes physical activity equivalent to walking 1.5 to 3 miles per day at 3 to 4 miles per hour, in addition to the light physical activity associated with typical day-to-day life.)

    A bar graph shows the estimated number of calories per day needed to maintain energy balance for various gender and age groups for moderately active lifestyles.

    Source: U.S.D.A.

    1. The mathematical model

      W=66x2+526x+1030

      describes the number of calories needed per day, W, by women in age group x with moderately active lifestyles. According to the model, how many calories per day are needed by women between the ages of 19 and 30, inclusive, with this lifestyle? Does this underestimate or overestimate the number shown by the graph? By how much?

    2. The mathematical model

      M=120x2+998x+590

      describes the number of calories needed per day, M, by men in age group x with moderately active lifestyles. According to the model, how many calories per day are needed by men between the ages of 19 and 30, inclusive, with this lifestyle? Does this underestimate or overestimate the number shown by the graph? By how much?

    3. Write a simplified rational expression that describes the ratio of the number of calories needed per day by women in age group x to the number of calories needed per day by men in age group x for people with moderately active lifestyles.

  4. 104. If three resistors with resistances R1, R2, and R3 are connected in parallel, their combined resistance is given by the expression

    An image shows a mathematical expression and a figure of three resistors with resistances R1, R2, and R3.

    Simplify the complex rational expression. Then find the combined resistance when R1 is 4 ohms, R2 is 8 ohms, and R3 is 12 ohms.

In Exercises 105–106, express the perimeter of each rectangle as a single rational expression.

  1. 105.

    A rectangle of length start fraction x over x + 3 end fraction and width start fraction x over x + 4 end fraction.
  2. 106.

    A rectangle of length start fraction x over x + 5 end fraction and width start fraction x over x + 6 end fraction.

Explaining the Concepts

  1. 107. What is a rational expression?

  2. 108. Explain how to determine which numbers must be excluded from the domain of a rational expression.

  3. 109. Explain how to simplify a rational expression.

  4. 110. Explain how to multiply rational expressions.

  5. 111. Explain how to divide rational expressions.

  6. 112. Explain how to add or subtract rational expressions with the same denominators.

  7. 113. Explain how to add rational expressions with different denominators. Use 3x+5+7(x+2)(x+5) in your explanation.

  8. 114. Explain how to find the least common denominator for denominators of x2100 and x220x+100.

  9. 115. Describe two ways to simplify 3x+2x21x2+2x.

Explain the error in Exercises 116118. Then rewrite the right side of the equation to correct the error that now exists.

  1. 116. 1a+1b=1a+b

  2. 117. 1x+7=1x+7

  3. 118. ax+ab=ax+b

Critical Thinking Exercises

Make Sense? In Exercises 119122, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 119. I evaluated 3x34x(x1) for x=1 and obtained 0.

  2. 120. The rational expressions

    714xand714+x

    can both be simplified by dividing each numerator and each denominator by 7.

  3. 121. When performing the division

    7xx+3÷(x+3)2x5,

    I began by dividing the numerator and the denominator by the common factor, x+3.

  4. 122. I subtracted 3x5x1 from x3x1 and obtained a constant.

In Exercises 123126, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 123. x225x5=x5

  2. 124. The expression 3y6y+2 simplifies to the consecutive integer that follows 4.

  3. 125. 2x1x7+3x1x75x2x7=0

  4. 126. 6+1x=7x

In Exercises 127129, perform the indicated operations.

  1. 127. 1xn11xn+11x2n1

  2. 128. (11x)(11x+1)(11x+2)(11x+3)

  3. 129. (xy)1+(xy)2

  4. 130. In one short sentence, five words or less, explain what

    1x+1x2+1x31x4+1x5+1x6

    does to each number x.

Preview Exercises

Exercises 131133 will help you prepare for the material covered in the next section.

  1. 131. If 6 is substituted for x in the equation

    2(x3)17=133(x+2),

    is the resulting statement true or false?

  2. 132. Multiply and simplify: 12 (x+24x13).

  3. 133. Evaluate

    bb24ac2a

    for a=2, b=9, and c =5.

P.6: Exercise Set

P.6 Exercise Set

Practice Exercises

In Exercises 16, find all numbers that must be excluded from the domain of each rational expression.

  1. 1. 7x3

  2. 2. 13x+9

  3. 3. x+5x225

  4. 4. x+7x249

  5. 5. x1x2+11x+10

  6. 6. x3x2+4x45

In Exercises 714, simplify each rational expression. Find all numbers that must be excluded from the domain of the simplified rational expression.

  1. 7. 3x9x26x+9

  2. 8. 4x8x24x+4

  3. 9. x212x+364x24

  4. 10. x28x+163x12

  5. 11. y2+7y18y23y+2

  6. 12. y24y5y2+5y+4

  7. 13. x2+12x+36x236

  8. 14. x214x+49x249

In Exercises 1532, multiply or divide as indicated.

  1. 15. x23x+92x+62x4

  2. 16. 6x+93x15x54x+6

  3. 17. x29x2x23xx2+x12

  4. 18. x24x24x+42x4x+2

  5. 19. x25x+6x22x3x21x24

  6. 20. x2+5x+6x2+x6x29x2x6

  7. 21. x38x24x+23x

  8. 22. x2+6x+9x3+271x+3

  9. 23. x+13÷3x+37

  10. 24. x+57÷4x+209

  11. 25. x24x÷x+2x2

  12. 26. x24x2÷x+24x8

  13. 27. 4x2+10x3÷6x2+15x29

  14. 28. x2+xx24÷x21x2+5x+6

  15. 29. x2252x2÷x2+10x+25x2+4x5

  16. 30. x24x2+3x10÷x2+5x+6x2+8x+15

  17. 31. x2+x12x2+x30x2+5x+6x22x3÷x+3x2+7x+6

  18. 32. x325x4x22x22x26x+5÷x2+5x7x+7

In Exercises 3368, add or subtract as indicated.

  1. 33. 4x+16x+5+8x+96x+5

  2. 34. 3x+23x+4+3x+63x+4

  3. 35. x22xx2+3x+x2+xx2+3x

  4. 36. x24xx2x6+4x4x2x6

  5. 37. 4x10x2x4x2

  6. 38. 2x+33x63x3x6

  7. 39. x2+3xx2+x12x212x2+x12

  8. 40. x24xx2x6x6x2x6

  9. 41. 5x+3

  10. 42. 74x

  11. 43. x4+5x6

  12. 44. 3x8+x12

  13. 45. 103x83

  14. 46. 7565x

  15. 47. 25xx+14x

  16. 48. 47x+x13x

  17. 49. 56x+x38x2

  18. 50. x+910x3+1115x2

  19. 51. 3x+4+6x+5

  20. 52. 8x2+2x3

  21. 53. 3x+13x

  22. 54. 4x3x+3

  23. 55. 2xx+2+x+2x2

  24. 56. 3xx3x+4x+2

  25. 57. x+5x5+x5x+5

  26. 58. x+3x3+x3x+3

  27. 59. 32x+4+23x+6

  28. 60. 52x+8+73x+12

  29. 61. 4x2+6x+9+4x+3

  30. 62. 35x+2+5x25x24

  31. 63. 3xx2+3x102xx2+x6

  32. 64. xx22x24xx27x+6

  33. 65. x+3x21x+2x1

  34. 66. x+5x24x+1x2

  35. 67. 4x2+x6x2+3x+23xx+1+5x+2

  36. 68. 6x2+17x40x2+x20+3x45xx+5

In Exercises 6982, simplify each complex rational expression.

  1. 69. x31x3

  2. 70. x41x4

  3. 71. 1+1x31x

  4. 72. 8+1x41x

  5. 73. 1x+1yx+y

  6. 74. 11xxy

  7. 75. xxx+3x+2

  8. 76. x3x3x2

  9. 77. 3x24x+27x24

  10. 78. xx2+13x24+1

  11. 79. 1x+11x22x3+1x3

  12. 80. 6x2+2x151x31x+5+1

  13. 81. 1(x+h)21x2h

  14. 82. x+hx+h+1xx+1h

Exercises 83–88 contain fractional expressions that occur frequently in calculus. Simplify each expression.

  1. 83. x13xx

  2. 84. x14xx

  3. 85. x2x2+2x2+2x2

  4. 86. 5x2+x25x25x2

  5. 87. 1x+h1xh

  6. 88. 1x+31x3

In Exercises 89–92, rationalize the numerator.

  1. 89. x+5x5

  2. 90. x+7x7

  3. 91. x+yx2y2

  4. 92. xyx2y2

Practice PLUS

In Exercises 93100, perform the indicated operations. Simplify the result, if possible.

  1. 93. (2x+3x+1x2+4x52x2+x3)2x+2

  2. 94. 1x22x8÷(1x41x+2)

  3. 95. (26x+1)(1+3x2)

  4. 96. (43x+2)(1+5x1)

  5. 97. y1(y+5)15

  6. 98. y1(y+2)12

  7. 99. (1a3b3ac+adbcbd1)cda2+ab+b2

  8. 100. aba2+ab+b2+(acadbc+bdacad+bcbd÷a3b3a3+b3)

Application Exercises

  1. 101. The rational expression

    130x100x

    describes the cost, in millions of dollars, to inoculate x percent of the population against a particular strain of flu.

    1. Evaluate the expression for x=40, x=80, and x=90. Describe the meaning of each evaluation in terms of percentage inoculated and cost.

    2. For what value of x is the expression undefined?

    3. What happens to the cost as x approaches 100%? How can you interpret this observation?

  2. 102. The average rate on a round-trip commute having a one-way distance d is given by the complex rational expression

    2ddr1+dr2,

    in which r1 and r2 are the average rates on the outgoing and return trips, respectively. Simplify the expression. Then find your average rate if you drive to campus averaging 40 miles per hour and return home on the same route averaging 30 miles per hour. Explain why the answer is not 35 miles per hour.

  3. 103. The bar graph shows the estimated number of calories per day needed to maintain energy balance for various gender and age groups for moderately active lifestyles. (Moderately active means a lifestyle that includes physical activity equivalent to walking 1.5 to 3 miles per day at 3 to 4 miles per hour, in addition to the light physical activity associated with typical day-to-day life.)

    A bar graph shows the estimated number of calories per day needed to maintain energy balance for various gender and age groups for moderately active lifestyles.

    Source: U.S.D.A.

    1. The mathematical model

      W=66x2+526x+1030

      describes the number of calories needed per day, W, by women in age group x with moderately active lifestyles. According to the model, how many calories per day are needed by women between the ages of 19 and 30, inclusive, with this lifestyle? Does this underestimate or overestimate the number shown by the graph? By how much?

    2. The mathematical model

      M=120x2+998x+590

      describes the number of calories needed per day, M, by men in age group x with moderately active lifestyles. According to the model, how many calories per day are needed by men between the ages of 19 and 30, inclusive, with this lifestyle? Does this underestimate or overestimate the number shown by the graph? By how much?

    3. Write a simplified rational expression that describes the ratio of the number of calories needed per day by women in age group x to the number of calories needed per day by men in age group x for people with moderately active lifestyles.

  4. 104. If three resistors with resistances R1, R2, and R3 are connected in parallel, their combined resistance is given by the expression

    An image shows a mathematical expression and a figure of three resistors with resistances R1, R2, and R3.

    Simplify the complex rational expression. Then find the combined resistance when R1 is 4 ohms, R2 is 8 ohms, and R3 is 12 ohms.

In Exercises 105–106, express the perimeter of each rectangle as a single rational expression.

  1. 105.

    A rectangle of length start fraction x over x + 3 end fraction and width start fraction x over x + 4 end fraction.
  2. 106.

    A rectangle of length start fraction x over x + 5 end fraction and width start fraction x over x + 6 end fraction.

Explaining the Concepts

  1. 107. What is a rational expression?

  2. 108. Explain how to determine which numbers must be excluded from the domain of a rational expression.

  3. 109. Explain how to simplify a rational expression.

  4. 110. Explain how to multiply rational expressions.

  5. 111. Explain how to divide rational expressions.

  6. 112. Explain how to add or subtract rational expressions with the same denominators.

  7. 113. Explain how to add rational expressions with different denominators. Use 3x+5+7(x+2)(x+5) in your explanation.

  8. 114. Explain how to find the least common denominator for denominators of x2100 and x220x+100.

  9. 115. Describe two ways to simplify 3x+2x21x2+2x.

Explain the error in Exercises 116118. Then rewrite the right side of the equation to correct the error that now exists.

  1. 116. 1a+1b=1a+b

  2. 117. 1x+7=1x+7

  3. 118. ax+ab=ax+b

Critical Thinking Exercises

Make Sense? In Exercises 119122, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 119. I evaluated 3x34x(x1) for x=1 and obtained 0.

  2. 120. The rational expressions

    714xand714+x

    can both be simplified by dividing each numerator and each denominator by 7.

  3. 121. When performing the division

    7xx+3÷(x+3)2x5,

    I began by dividing the numerator and the denominator by the common factor, x+3.

  4. 122. I subtracted 3x5x1 from x3x1 and obtained a constant.

In Exercises 123126, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 123. x225x5=x5

  2. 124. The expression 3y6y+2 simplifies to the consecutive integer that follows 4.

  3. 125. 2x1x7+3x1x75x2x7=0

  4. 126. 6+1x=7x

In Exercises 127129, perform the indicated operations.

  1. 127. 1xn11xn+11x2n1

  2. 128. (11x)(11x+1)(11x+2)(11x+3)

  3. 129. (xy)1+(xy)2

  4. 130. In one short sentence, five words or less, explain what

    1x+1x2+1x31x4+1x5+1x6

    does to each number x.

Preview Exercises

Exercises 131133 will help you prepare for the material covered in the next section.

  1. 131. If 6 is substituted for x in the equation

    2(x3)17=133(x+2),

    is the resulting statement true or false?

  2. 132. Multiply and simplify: 12 (x+24x13).

  3. 133. Evaluate

    bb24ac2a

    for a=2, b=9, and c =5.

P.6: Exercise Set

P.6 Exercise Set

Practice Exercises

In Exercises 16, find all numbers that must be excluded from the domain of each rational expression.

  1. 1. 7x3

  2. 2. 13x+9

  3. 3. x+5x225

  4. 4. x+7x249

  5. 5. x1x2+11x+10

  6. 6. x3x2+4x45

In Exercises 714, simplify each rational expression. Find all numbers that must be excluded from the domain of the simplified rational expression.

  1. 7. 3x9x26x+9

  2. 8. 4x8x24x+4

  3. 9. x212x+364x24

  4. 10. x28x+163x12

  5. 11. y2+7y18y23y+2

  6. 12. y24y5y2+5y+4

  7. 13. x2+12x+36x236

  8. 14. x214x+49x249

In Exercises 1532, multiply or divide as indicated.

  1. 15. x23x+92x+62x4

  2. 16. 6x+93x15x54x+6

  3. 17. x29x2x23xx2+x12

  4. 18. x24x24x+42x4x+2

  5. 19. x25x+6x22x3x21x24

  6. 20. x2+5x+6x2+x6x29x2x6

  7. 21. x38x24x+23x

  8. 22. x2+6x+9x3+271x+3

  9. 23. x+13÷3x+37

  10. 24. x+57÷4x+209

  11. 25. x24x÷x+2x2

  12. 26. x24x2÷x+24x8

  13. 27. 4x2+10x3÷6x2+15x29

  14. 28. x2+xx24÷x21x2+5x+6

  15. 29. x2252x2÷x2+10x+25x2+4x5

  16. 30. x24x2+3x10÷x2+5x+6x2+8x+15

  17. 31. x2+x12x2+x30x2+5x+6x22x3÷x+3x2+7x+6

  18. 32. x325x4x22x22x26x+5÷x2+5x7x+7

In Exercises 3368, add or subtract as indicated.

  1. 33. 4x+16x+5+8x+96x+5

  2. 34. 3x+23x+4+3x+63x+4

  3. 35. x22xx2+3x+x2+xx2+3x

  4. 36. x24xx2x6+4x4x2x6

  5. 37. 4x10x2x4x2

  6. 38. 2x+33x63x3x6

  7. 39. x2+3xx2+x12x212x2+x12

  8. 40. x24xx2x6x6x2x6

  9. 41. 5x+3

  10. 42. 74x

  11. 43. x4+5x6

  12. 44. 3x8+x12

  13. 45. 103x83

  14. 46. 7565x

  15. 47. 25xx+14x

  16. 48. 47x+x13x

  17. 49. 56x+x38x2

  18. 50. x+910x3+1115x2

  19. 51. 3x+4+6x+5

  20. 52. 8x2+2x3

  21. 53. 3x+13x

  22. 54. 4x3x+3

  23. 55. 2xx+2+x+2x2

  24. 56. 3xx3x+4x+2

  25. 57. x+5x5+x5x+5

  26. 58. x+3x3+x3x+3

  27. 59. 32x+4+23x+6

  28. 60. 52x+8+73x+12

  29. 61. 4x2+6x+9+4x+3

  30. 62. 35x+2+5x25x24

  31. 63. 3xx2+3x102xx2+x6

  32. 64. xx22x24xx27x+6

  33. 65. x+3x21x+2x1

  34. 66. x+5x24x+1x2

  35. 67. 4x2+x6x2+3x+23xx+1+5x+2

  36. 68. 6x2+17x40x2+x20+3x45xx+5

In Exercises 6982, simplify each complex rational expression.

  1. 69. x31x3

  2. 70. x41x4

  3. 71. 1+1x31x

  4. 72. 8+1x41x

  5. 73. 1x+1yx+y

  6. 74. 11xxy

  7. 75. xxx+3x+2

  8. 76. x3x3x2

  9. 77. 3x24x+27x24

  10. 78. xx2+13x24+1

  11. 79. 1x+11x22x3+1x3

  12. 80. 6x2+2x151x31x+5+1

  13. 81. 1(x+h)21x2h

  14. 82. x+hx+h+1xx+1h

Exercises 83–88 contain fractional expressions that occur frequently in calculus. Simplify each expression.

  1. 83. x13xx

  2. 84. x14xx

  3. 85. x2x2+2x2+2x2

  4. 86. 5x2+x25x25x2

  5. 87. 1x+h1xh

  6. 88. 1x+31x3

In Exercises 89–92, rationalize the numerator.

  1. 89. x+5x5

  2. 90. x+7x7

  3. 91. x+yx2y2

  4. 92. xyx2y2

Practice PLUS

In Exercises 93100, perform the indicated operations. Simplify the result, if possible.

  1. 93. (2x+3x+1x2+4x52x2+x3)2x+2

  2. 94. 1x22x8÷(1x41x+2)

  3. 95. (26x+1)(1+3x2)

  4. 96. (43x+2)(1+5x1)

  5. 97. y1(y+5)15

  6. 98. y1(y+2)12

  7. 99. (1a3b3ac+adbcbd1)cda2+ab+b2

  8. 100. aba2+ab+b2+(acadbc+bdacad+bcbd÷a3b3a3+b3)

Application Exercises

  1. 101. The rational expression

    130x100x

    describes the cost, in millions of dollars, to inoculate x percent of the population against a particular strain of flu.

    1. Evaluate the expression for x=40, x=80, and x=90. Describe the meaning of each evaluation in terms of percentage inoculated and cost.

    2. For what value of x is the expression undefined?

    3. What happens to the cost as x approaches 100%? How can you interpret this observation?

  2. 102. The average rate on a round-trip commute having a one-way distance d is given by the complex rational expression

    2ddr1+dr2,

    in which r1 and r2 are the average rates on the outgoing and return trips, respectively. Simplify the expression. Then find your average rate if you drive to campus averaging 40 miles per hour and return home on the same route averaging 30 miles per hour. Explain why the answer is not 35 miles per hour.

  3. 103. The bar graph shows the estimated number of calories per day needed to maintain energy balance for various gender and age groups for moderately active lifestyles. (Moderately active means a lifestyle that includes physical activity equivalent to walking 1.5 to 3 miles per day at 3 to 4 miles per hour, in addition to the light physical activity associated with typical day-to-day life.)

    A bar graph shows the estimated number of calories per day needed to maintain energy balance for various gender and age groups for moderately active lifestyles.

    Source: U.S.D.A.

    1. The mathematical model

      W=66x2+526x+1030

      describes the number of calories needed per day, W, by women in age group x with moderately active lifestyles. According to the model, how many calories per day are needed by women between the ages of 19 and 30, inclusive, with this lifestyle? Does this underestimate or overestimate the number shown by the graph? By how much?

    2. The mathematical model

      M=120x2+998x+590

      describes the number of calories needed per day, M, by men in age group x with moderately active lifestyles. According to the model, how many calories per day are needed by men between the ages of 19 and 30, inclusive, with this lifestyle? Does this underestimate or overestimate the number shown by the graph? By how much?

    3. Write a simplified rational expression that describes the ratio of the number of calories needed per day by women in age group x to the number of calories needed per day by men in age group x for people with moderately active lifestyles.

  4. 104. If three resistors with resistances R1, R2, and R3 are connected in parallel, their combined resistance is given by the expression

    An image shows a mathematical expression and a figure of three resistors with resistances R1, R2, and R3.

    Simplify the complex rational expression. Then find the combined resistance when R1 is 4 ohms, R2 is 8 ohms, and R3 is 12 ohms.

In Exercises 105–106, express the perimeter of each rectangle as a single rational expression.

  1. 105.

    A rectangle of length start fraction x over x + 3 end fraction and width start fraction x over x + 4 end fraction.
  2. 106.

    A rectangle of length start fraction x over x + 5 end fraction and width start fraction x over x + 6 end fraction.

Explaining the Concepts

  1. 107. What is a rational expression?

  2. 108. Explain how to determine which numbers must be excluded from the domain of a rational expression.

  3. 109. Explain how to simplify a rational expression.

  4. 110. Explain how to multiply rational expressions.

  5. 111. Explain how to divide rational expressions.

  6. 112. Explain how to add or subtract rational expressions with the same denominators.

  7. 113. Explain how to add rational expressions with different denominators. Use 3x+5+7(x+2)(x+5) in your explanation.

  8. 114. Explain how to find the least common denominator for denominators of x2100 and x220x+100.

  9. 115. Describe two ways to simplify 3x+2x21x2+2x.

Explain the error in Exercises 116118. Then rewrite the right side of the equation to correct the error that now exists.

  1. 116. 1a+1b=1a+b

  2. 117. 1x+7=1x+7

  3. 118. ax+ab=ax+b

Critical Thinking Exercises

Make Sense? In Exercises 119122, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 119. I evaluated 3x34x(x1) for x=1 and obtained 0.

  2. 120. The rational expressions

    714xand714+x

    can both be simplified by dividing each numerator and each denominator by 7.

  3. 121. When performing the division

    7xx+3÷(x+3)2x5,

    I began by dividing the numerator and the denominator by the common factor, x+3.

  4. 122. I subtracted 3x5x1 from x3x1 and obtained a constant.

In Exercises 123126, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 123. x225x5=x5

  2. 124. The expression 3y6y+2 simplifies to the consecutive integer that follows 4.

  3. 125. 2x1x7+3x1x75x2x7=0

  4. 126. 6+1x=7x

In Exercises 127129, perform the indicated operations.

  1. 127. 1xn11xn+11x2n1

  2. 128. (11x)(11x+1)(11x+2)(11x+3)

  3. 129. (xy)1+(xy)2

  4. 130. In one short sentence, five words or less, explain what

    1x+1x2+1x31x4+1x5+1x6

    does to each number x.

Preview Exercises

Exercises 131133 will help you prepare for the material covered in the next section.

  1. 131. If 6 is substituted for x in the equation

    2(x3)17=133(x+2),

    is the resulting statement true or false?

  2. 132. Multiply and simplify: 12 (x+24x13).

  3. 133. Evaluate

    bb24ac2a

    for a=2, b=9, and c =5.

Section P.7: Equations

Section P.7 Equations

Learning Objectives

What you’ll learn

  1. 1 Solve linear equations in one variable.

  2. 2 Solve linear equations containing fractions.

  3. 3 Solve rational equations with variables in the denominators.

  4. 4 Solve a formula for a variable.

  5. 5 Solve equations involving absolute value.

  6. 6 Solve quadratic equations by factoring.

  7. 7 Solve quadratic equations by the square root property.

  8. 8 Solve quadratic equations by completing the square.

  9. 9 Solve quadratic equations using the quadratic formula.

  10. 10 Use the discriminant to determine the number and type of solutions of quadratic equations.

  11. 11 Determine the most efficient method to use when solving a quadratic equation.

  12. 12 Solve radical equations.

I’m very well acquainted, too, with matters mathematical, I understand equations, both simple and quadratical. About binomial theorem I’m teeming with a lot of news, With many cheerful facts about the square of the hypotenuse.

—Gilbert and Sullivan, The Pirates of Penzance

Equations quadratical? Cheerful news about the square of the hypotenuse? You’ve come to the right place. In this section, we will review how to solve a variety of equations, including linear equations, quadratic equations, and radical equations. (Yes, it’s quadratic and not quadratical, despite the latter’s rhyme with mathematical.) In the next section, we will look at applications of quadratic equations, introducing (cheerfully, of course) the Pythagorean Theorem and the square of the hypotenuse.

Objective 1: Solve linear equations in one variable

Solving Linear Equations in One Variable

  1. Objective 1Solve linear equations in one variable.

Watch Video

We begin with a general definition of a linear equation in one variable.

Definition of a Linear Equation

A linear equation in one variable x is an equation that can be written in the form

ax+b=0,

where a and b are real numbers, and a0.

An example of a linear equation in one variable is

4x+12=0.

Solving an equation in x involves determining all values of x that result in a true statement when substituted into the equation. Such values are solutions, or roots, of the equation. For example, substitute 3 for x in 4x+12=0. We obtain

4(3)+12=0,or12+12=0.

This simplifies to the true statement 0=0. Thus, 3 is a solution of the equation 4x+12=0. We also say that 3 satisfies the equation 4x+12=0, because when we substitute 3 for x, a true statement results. The set of all such solutions is called the equation’s solution set. For example, the solution set of the equation 4x+12=0 is {3} because 3 is the equation’s only solution.

Two or more equations that have the same solution set are called equivalent equations. For example, the equations

4x+12=0and4x=12andx=3

are equivalent equations because the solution set for each is {3}. To solve a linear equation in x, we transform the equation into an equivalent equation one or more times. Our final equivalent equation should be of the form

x=a number.

The solution set of this equation is the set consisting of the number.

To generate equivalent equations, we will use the principles in the box.

Generating Equivalent Equations

An equation can be transformed into an equivalent equation by one or more of the following operations:

A table titled, Generating equivalent equations with corresponding examples.

If you look closely at the equations in the box, you will notice that we have solved the equation 3(x6)=6xx. The final equation, x=9, with x isolated on the left side, shows that {9} is the solution set. The idea in solving a linear equation is to get the variable by itself on one side of the equal sign and a number by itself on the other side.

Here is a step-by-step procedure for solving a linear equation in one variable. Not all of these steps are necessary to solve every equation.

Solving a Linear Equation

  1. Simplify the algebraic expression on each side by removing grouping symbols and combining like terms.

  2. Collect all the variable terms on one side and all the numbers, or constant terms, on the other side.

  3. Isolate the variable and solve.

  4. Check the proposed solution in the original equation.

Example 1 Solving a Linear Equation

Solve and check: 2(x3)17=133(x+2).

Solution

  1. Step 1 SIMPLIFY THE ALGEBRAIC EXPRESSION ON EACH SIDE.

    An image shows step 1 simplifies the algebraic expression on each side.

  2. Step 2 COLLECT VARIABLE TERMS ON ONE SIDE AND CONSTANT TERMS ON THE OTHER SIDE. We will collect variable terms of 2x23=3x+7 on the left by adding 3x to both sides. We will collect the numbers on the right by adding 23 to both sides.

    2x23=3x+7This is the equation with each sidesimplified.2x23+3x=3x+7+3xAdd 3x  to both sides.5x23=7Simplify: 2x+3x=5x.5x23+23=7+23Add 23 to both sides.5x=30Simplify.
  3. Step 3. ISOLATE THE VARIABLE AND SOLVE. We isolate the variable, x, by dividing both sides of 5x=30 by 5.

    5x5=305Divide both sides by 5.x=6Simplify.
  4. Step 4. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL EQUATION. Substitute 6 for x in the original equation.

    2(x3)17=133(x+2)This is the original equation.2(63)17=?133(6+2)Substitute 6 for x.2(3)17=?133(8)Simplify inside parentheses.617=?1324Multiply.11=11Subtract.

    The true statement 11=11 verifies that the solution set is {6}.

Check Point 1

  • Solve and check: 4(2x+1)=29+3(2x5).

Objective 1: Solve linear equations in one variable

Solving Linear Equations in One Variable

  1. Objective 1Solve linear equations in one variable.

Watch Video

We begin with a general definition of a linear equation in one variable.

Definition of a Linear Equation

A linear equation in one variable x is an equation that can be written in the form

ax+b=0,

where a and b are real numbers, and a0.

An example of a linear equation in one variable is

4x+12=0.

Solving an equation in x involves determining all values of x that result in a true statement when substituted into the equation. Such values are solutions, or roots, of the equation. For example, substitute 3 for x in 4x+12=0. We obtain

4(3)+12=0,or12+12=0.

This simplifies to the true statement 0=0. Thus, 3 is a solution of the equation 4x+12=0. We also say that 3 satisfies the equation 4x+12=0, because when we substitute 3 for x, a true statement results. The set of all such solutions is called the equation’s solution set. For example, the solution set of the equation 4x+12=0 is {3} because 3 is the equation’s only solution.

Two or more equations that have the same solution set are called equivalent equations. For example, the equations

4x+12=0and4x=12andx=3

are equivalent equations because the solution set for each is {3}. To solve a linear equation in x, we transform the equation into an equivalent equation one or more times. Our final equivalent equation should be of the form

x=a number.

The solution set of this equation is the set consisting of the number.

To generate equivalent equations, we will use the principles in the box.

Generating Equivalent Equations

An equation can be transformed into an equivalent equation by one or more of the following operations:

A table titled, Generating equivalent equations with corresponding examples.

If you look closely at the equations in the box, you will notice that we have solved the equation 3(x6)=6xx. The final equation, x=9, with x isolated on the left side, shows that {9} is the solution set. The idea in solving a linear equation is to get the variable by itself on one side of the equal sign and a number by itself on the other side.

Here is a step-by-step procedure for solving a linear equation in one variable. Not all of these steps are necessary to solve every equation.

Solving a Linear Equation

  1. Simplify the algebraic expression on each side by removing grouping symbols and combining like terms.

  2. Collect all the variable terms on one side and all the numbers, or constant terms, on the other side.

  3. Isolate the variable and solve.

  4. Check the proposed solution in the original equation.

Example 1 Solving a Linear Equation

Solve and check: 2(x3)17=133(x+2).

Solution

  1. Step 1 SIMPLIFY THE ALGEBRAIC EXPRESSION ON EACH SIDE.

    An image shows step 1 simplifies the algebraic expression on each side.

  2. Step 2 COLLECT VARIABLE TERMS ON ONE SIDE AND CONSTANT TERMS ON THE OTHER SIDE. We will collect variable terms of 2x23=3x+7 on the left by adding 3x to both sides. We will collect the numbers on the right by adding 23 to both sides.

    2x23=3x+7This is the equation with each sidesimplified.2x23+3x=3x+7+3xAdd 3x  to both sides.5x23=7Simplify: 2x+3x=5x.5x23+23=7+23Add 23 to both sides.5x=30Simplify.
  3. Step 3. ISOLATE THE VARIABLE AND SOLVE. We isolate the variable, x, by dividing both sides of 5x=30 by 5.

    5x5=305Divide both sides by 5.x=6Simplify.
  4. Step 4. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL EQUATION. Substitute 6 for x in the original equation.

    2(x3)17=133(x+2)This is the original equation.2(63)17=?133(6+2)Substitute 6 for x.2(3)17=?133(8)Simplify inside parentheses.617=?1324Multiply.11=11Subtract.

    The true statement 11=11 verifies that the solution set is {6}.

Check Point 1

  • Solve and check: 4(2x+1)=29+3(2x5).

Objective 2: Solve linear equations containing fractions

Linear Equations with Fractions

  1. Objective 2Solve linear equations containing fractions.

Watch Video

Equations are easier to solve when they do not contain fractions. How do we remove fractions from an equation? We begin by multiplying both sides of the equation by the least common denominator of any fractions in the equation. The least common denominator is the smallest number that all denominators will divide into. Multiplying every term on both sides of the equation by the least common denominator will eliminate the fractions in the equation. Example 2 shows how we “clear an equation of fractions.”

Example 2 Solving a Linear Equation Involving Fractions

Solve and check: x+24x13=2.

Solution

The fractional terms have denominators of 4 and 3. The smallest number that is divisible by 4 and 3 is 12. We begin by multiplying both sides of the equation by 12, the least common denominator.

An image shows the steps of solving a linear equation involving fractions.

Isolate x by multiplying or dividing both sides of this equation by 1.

x1=141Divide both sides by  1.x=14Simplify.

Check the proposed solution. Substitute 14 for x in the original equation. You should obtain 2=2. This true statement verifies that the solution set is {14}.

Check Point 2

  • Solve and check: x34=514x+57.

Objective 3: Solve rational equations with variables in the denominators

Rational Equations

  1. Objective 3Solve rational equations with variables in the denominators.

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A rational equation is an equation containing one or more rational expressions. In Example 2, we solved a rational equation with constants in the denominators. This rational equation was a linear equation. Now, let’s consider a rational equation such as

3x+6+1x2=4x2+4x12.

Can you see how this rational equation differs from the rational equation that we solved earlier? The variable appears in the denominators. Although this rational equation is not a linear equation, the solution procedure still involves multiplying each side by the least common denominator. However, we must avoid any values of the variable that make a denominator zero.

Example 3 Solving a Rational Equation

Solve: 3x+6+1x2=4x2+4x12.

Solution

To identify values of x that make denominators zero, let’s factor x2+4x12, the denominator on the right. This factorization is also necessary in identifying the least common denominator.

Start fraction 3 over x + 6 end fraction + start fraction 1 over x minus 2 end fraction = start fraction 4 over left parenthesis x + 6 right parenthesis, left parenthesis x minus 2 right parenthesis end fraction.

We see that x cannot equal 6 or 2. The least common denominator is (x+6)(x2).

The image shows mathematical expressions to solve a rational equation to find the value of x.

Check the proposed solution. Substitute 1 for x in the original equation. You should obtain 47=47. This true statement verifies that the solution set is {1}.

Check Point 3

  • Solve: 6x+35x2=20x2+x6.

Example 4 Solving a Rational Equation

Solve: 1x+1=2x211x1.

Solution

We begin by factoring x21.

Start fraction 1 over x + 1 end fraction = start fraction 2 over left parenthesis x + 1 right parenthesis, left parenthesis x minus 1 right parenthesis end fraction minus start fraction 1 over x minus 1 end fraction.

We see that x cannot equal 1 or 1. The least common denominator is (x+1)(x1).

The image shows mathematical expressions to solve a rational equation to find the value of x.

The proposed solution, 1, is not a solution because of the restriction that x1. There is no solution to this equation. The solution set for this equation contains no elements. The solution set is Ø, the empty set.

Check Point 4

  • Solve: 1x+2=4x241x2.

Objective 4: Solve a formula for a variable

Solving a Formula for One of Its Variables

  1. Objective 4Solve a formula for a variable.

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Solving a formula for a variable means rewriting the formula so that the variable is isolated on one side of the equation. It does not mean obtaining a numerical value for that variable.

To solve a formula for one of its variables, treat that variable as if it were the only variable in the equation. Think of the other variables as if they were numbers.

Example 5 Solving a Formula for a Variable

If you wear glasses, did you know that each lens has a measurement called its focal length, f? When an object is in focus, its distance from the lens, p, and the distance from the lens to your retina, q, satisfy the formula

1p+1q=1f.

(See Figure P.12.) Solve this formula for p.

Figure P.12

An image shows an eye wearing a glass which shows that each lens has a measurement called its focal length f and when an object is in focus, its distance from the lens, p, and the distance from the lens to your retina, q.

Solution

Our goal is to isolate the variable p. We begin by multiplying both sides by the least common denominator, pqf, to clear the equation of fractions.

The image shows a mathematical expression to solve a formula for a variable to find the value of p.

To collect terms with p on one side of qf+pf=pq, subtract pf from both sides. Then factor p from the two resulting terms on the right to convert two occurrences of p into one.

qf+pf=pqThis is the equation cleared of fractions.qf+pfpf=pqpfSubtract pf from both sides.qf=pqpfSimplify.qf=p(qf)Factor out p,the specified variable.qfqf=p(qf)qfDivide both sides by qf and solve for p.qfqf=pSimplify.

Check Point 5

  • Solve for q:1p+1q=1f.

Objective 5: Solve equations involving absolute value

Equations Involving Absolute Value

  1. Objective 5Solve equations involving absolute value.

Watch Video

We have seen that the absolute value of x, denoted |x|, describes the distance of x from zero on a number line. Now consider an absolute value equation, such as

|x|=2.

This means that we must determine real numbers whose distance from the origin on a number line is 2. Figure P.13 shows that there are two numbers such that |x|=2, namely, 2 and 2. We write x=2 or x=2. This observation can be generalized as follows:

Figure P.13

A number line.
Figure P.13 Full Alternative Text

Rewriting an Absolute Value Equation without Absolute Value Bars

If c is a positive real number and u represents any algebraic expression, then |u|=c is equivalent to u=c or u=c.

Example 6 Solving an Equation Involving Absolute Value

Solve: 5|14x|15=0.

Solution

Solving an equation 5 vertical bar 1 minus 4 x vertical bar minus 15 = 0 in six steps.

Take a moment to check 12 and 1, the proposed solutions, in the original equation, 5|14x|15=0. In each case, you should obtain the true statement 0=0. The solution set is {12,1}.

Check Point 6

  • Solve: 4|12x|20=0.

The absolute value of a number is never negative. Thus, if u is an algebraic expression and c is a negative number, then |u|=c has no solution. For example, the equation |3x6|=2 has no solution because |3x6| cannot be negative. The solution set is Ø, the empty set.

The absolute value of 0 is 0. Thus, if u is an algebraic expression and |u|=0, the solution is found by solving u=0. For example, the solution of |x2|=0 is obtained by solving x2=0. The solution is 2 and the solution set is {2}.

Objective 6: Solve quadratic equations by factoring

Quadratic Equations and Factoring

  1. Objective 6Solve quadratic equations by factoring.

Watch Video

Linear equations are first-degree polynomial equations of the form ax+b=0. Quadratic equations are second-degree polynomial equations and contain an additional term involving the square of the variable.

Definition of a Quadratic Equation

A quadratic equation in x is an equation that can be written in the standard form

ax2+bx+c=0,

where a, b, and c are real numbers, with a0. A quadratic equation in x is also called a second-degree polynomial equation in x.

Here are examples of quadratic equations in standard form:

4 x squared minus 2 x = 0. Text pointing at 4 x squared reads, a = 4. Text pointing at negative 2 x reads, b = negative 2. Text pointing at = sign reads, c = 0.
P.8-170 Full Alternative Text

Some quadratic equations, including the two shown above, can be solved by factoring and using the zero-product principle.

The Zero-Product Principle

If the product of two algebraic expressions is zero, then at least one of the factors is equal to zero.

If AB=0, then A=0 or B=0.

The zero-product principle can be applied only when a quadratic equation is in standard form, with zero on one side of the equation.

Solving a Quadratic Equation by Factoring

  1. If necessary, rewrite the equation in the standard form ax2+bx+c=0, moving all nonzero terms to one side, thereby obtaining zero on the other side.

  2. Factor completely.

  3. Apply the zero-product principle, setting each factor containing a variable equal to zero.

  4. Solve the equations in step 3.

  5. Check the solutions in the original equation.

Example 7 Solving Quadratic Equations by Factoring

Solve by factoring:

  1. 4x22x=0

  2. 2x2+7x=4.

Solution

  1. We begin with 4x22x=0.

    1. Step 1 MOVE ALL NONZERO TERMS TO ONE SIDE AND OBTAIN ZERO ON THE OTHER SIDE. All nonzero terms are already on the left and zero is on the other side, so we can skip this step.

    2. Step 2 FACTOR. We factor out 2x from the two terms on the left side.

      4x22x=0This is the given equation.2x(2x1)=0Factor.
    3. Steps 3 and 4 SET EACH FACTOR EQUAL TO ZERO AND SOLVE THE RESULTING EQUATIONS. We apply the zero-product principle to 2x(2x1)=0.

      2x=0or2x1=0.2x=0or12x=0.2x=0or12x=12
    4. Step 5 CHECK THE SOLUTIONS IN THE ORIGINAL EQUATION.

      Check 0:

      Check 12:

      4x22x=0.40220=?0.00=?0.0=0,true
      4x22x=0.4(12)22(12)=?0.4(14)2(12)=?0.11=?0.0=0,true

      The solution set is {0,12}

  2. Next, we solve 2x2+7x=4.

    1. Step 1 MOVE ALL NONZERO TERMS TO ONE SIDE AND OBTAIN ZERO ON THE OTHER SIDE. Subtract 4 from both sides and write the equation in standard form.

      2x2+7x=4This is the given equation.2x2+7x4=44Subtract 4 from both sides.2x2+7x4=0Simplify.
    2. Step 2 FACTOR.

      2x2+7x4=0(2x1)(x+4)=0

    3. Steps 3 and 4 SET EACH FACTOR EQUAL TO ZERO AND SOLVE THE RESULTING EQUATIONS.

      2x1=0orx+4=02x=1x=4x=12
    4. Step 5 CHECK THE SOLUTIONS IN THE ORIGINAL EQUATION.

      Check: 12

      Check: 4:

      2x2+7x=4,2(12)2+7(12)=?4,12+72=?4,4=4,true
      2x2+7x=4,2(4)2+7(4)=?4,32+(28)=?4,4=4,true

      The solution set is {4,12}.

Check Point 7

  • Solve by factoring:

    1. 3x2=9x

    2. 2x2=1x.

Objective 6: Solve quadratic equations by factoring

Quadratic Equations and Factoring

  1. Objective 6Solve quadratic equations by factoring.

Watch Video

Linear equations are first-degree polynomial equations of the form ax+b=0. Quadratic equations are second-degree polynomial equations and contain an additional term involving the square of the variable.

Definition of a Quadratic Equation

A quadratic equation in x is an equation that can be written in the standard form

ax2+bx+c=0,

where a, b, and c are real numbers, with a0. A quadratic equation in x is also called a second-degree polynomial equation in x.

Here are examples of quadratic equations in standard form:

4 x squared minus 2 x = 0. Text pointing at 4 x squared reads, a = 4. Text pointing at negative 2 x reads, b = negative 2. Text pointing at = sign reads, c = 0.
P.8-170 Full Alternative Text

Some quadratic equations, including the two shown above, can be solved by factoring and using the zero-product principle.

The Zero-Product Principle

If the product of two algebraic expressions is zero, then at least one of the factors is equal to zero.

If AB=0, then A=0 or B=0.

The zero-product principle can be applied only when a quadratic equation is in standard form, with zero on one side of the equation.

Solving a Quadratic Equation by Factoring

  1. If necessary, rewrite the equation in the standard form ax2+bx+c=0, moving all nonzero terms to one side, thereby obtaining zero on the other side.

  2. Factor completely.

  3. Apply the zero-product principle, setting each factor containing a variable equal to zero.

  4. Solve the equations in step 3.

  5. Check the solutions in the original equation.

Example 7 Solving Quadratic Equations by Factoring

Solve by factoring:

  1. 4x22x=0

  2. 2x2+7x=4.

Solution

  1. We begin with 4x22x=0.

    1. Step 1 MOVE ALL NONZERO TERMS TO ONE SIDE AND OBTAIN ZERO ON THE OTHER SIDE. All nonzero terms are already on the left and zero is on the other side, so we can skip this step.

    2. Step 2 FACTOR. We factor out 2x from the two terms on the left side.

      4x22x=0This is the given equation.2x(2x1)=0Factor.
    3. Steps 3 and 4 SET EACH FACTOR EQUAL TO ZERO AND SOLVE THE RESULTING EQUATIONS. We apply the zero-product principle to 2x(2x1)=0.

      2x=0or2x1=0.2x=0or12x=0.2x=0or12x=12
    4. Step 5 CHECK THE SOLUTIONS IN THE ORIGINAL EQUATION.

      Check 0:

      Check 12:

      4x22x=0.40220=?0.00=?0.0=0,true
      4x22x=0.4(12)22(12)=?0.4(14)2(12)=?0.11=?0.0=0,true

      The solution set is {0,12}

  2. Next, we solve 2x2+7x=4.

    1. Step 1 MOVE ALL NONZERO TERMS TO ONE SIDE AND OBTAIN ZERO ON THE OTHER SIDE. Subtract 4 from both sides and write the equation in standard form.

      2x2+7x=4This is the given equation.2x2+7x4=44Subtract 4 from both sides.2x2+7x4=0Simplify.
    2. Step 2 FACTOR.

      2x2+7x4=0(2x1)(x+4)=0

    3. Steps 3 and 4 SET EACH FACTOR EQUAL TO ZERO AND SOLVE THE RESULTING EQUATIONS.

      2x1=0orx+4=02x=1x=4x=12
    4. Step 5 CHECK THE SOLUTIONS IN THE ORIGINAL EQUATION.

      Check: 12

      Check: 4:

      2x2+7x=4,2(12)2+7(12)=?4,12+72=?4,4=4,true
      2x2+7x=4,2(4)2+7(4)=?4,32+(28)=?4,4=4,true

      The solution set is {4,12}.

Check Point 7

  • Solve by factoring:

    1. 3x2=9x

    2. 2x2=1x.

Objective 7: Solve quadratic equations by the square root property

Solving Quadratic Equations by the Square Root Property

  1. Objective 7Solve quadratic equations by the square root property.

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Quadratic equations of the form u2=d, where u is an algebraic expression and d is a nonzero real number, can be solved by the square root property. First, isolate the squared expression u2 on one side of the equation and the number d on the other side. Then take the square root of both sides. Remember, there are two numbers whose square is d. One number is d and one is d.

We can use factoring to verify that u2=d has these two solutions.

u2=daThis is the given equation.u2d=0aMove all terms to one side and obtain zero on the other side.(u+d)(ud)=0aFactor.u+d=0orud=0aSet each factor equal to zero.u=dordu=dSolve the resulting equations.

Because the solutions differ only in sign, we can write them in abbreviated notation as u=±d. We read this as “u equals positive or negative the square root of d” or “u equals plus or minus the square root of d.”

Now that we have verified these solutions, we can solve u2=d directly by taking square roots. This process is called the square root property.

The Square Root Property

If u is an algebraic expression and d is a nonzero number, then u2=d has exactly two solutions:

Ifu2=d, thenu=d oru=d.

Equivalently,

Ifu2=d, thenu=±d.

Before you can apply the square root property, a squared expression must be isolated on one side of the equation.

Example 8 Solving Quadratic Equations by the Square Root Property

Solve by the square root property:

  1. 3x215=0

  2. (x2)2=6.

Solution

To apply the square root property, we need a squared expression by itself on one side of the equation.

3 x squared minus 15 = 0. Text pointing at x squared reads, we want x squared by itself. Left parenthesis x minus 2 right parenthesis squared = 6. Text pointing at minus sign reads The squared expression is by itself.
  1.  

    3x215=0This is the original equation.3x2=15Add 15 to both sides.x2=5Divide both sides by 3.x=5 or x=5Apply the square root property. Equivalently,x=±5.

    By checking both proposed solutions in the original equation, we can confirm that the solution set is {5,5} or {±5}.

  2.  

    (x2)2=6This is the original equation.x2=±6Apply the square root property.x=2±6Add 2 to both sides.

    By checking both values in the original equation, we can confirm that the solution set is {2+6, 26} or {2±6}.

Check Point 8

  • Solve by the square root property:

    1. 3x221=0

    2. (x+5)2=11.

Objective 8: Solve quadratic equations by completing the square

Solving Quadratic Equations by Completing the Square

  1. Objective 8Solve quadratic equations by completing the square.

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How do we solve an equation in the form ax2+bx+c=0 if the trinomial ax2+bx+c cannot be factored? We cannot use the zero-product principle in such a case. However, we can convert the equation into an equivalent equation that can be solved using the square root property. This is accomplished by completing the square.

Completing the Square

If x2+bx is a binomial, then by adding (b2)2, which is the square of half the coefficient of x, a perfect square trinomial will result. That is,

x2+bx+(b2)2=(x+b2)2.

We can solve any quadratic equation by completing the square and then applying the square root property.

Solving a Quadratic Equation by Completing the Square

To solve ax2+bx+c=0 by completing the square:

  1. If a, the leading coefficient, is not 1, divide both sides by a. This makes the coefficient of the x2-term 1.

  2. Isolate the variable terms on one side of the equation and the constant term on the other side of the equation.

  3. Complete the square.

    1. Add the square of half the coefficient of x to both sides of the equation.

    2. Factor the resulting perfect square trinomial.

  4. Use the square root property and solve for x.

Example 9 Solving a Quadratic Equation by Completing the Square

Solve by completing the square: x26x+4=0.

Solution

  1. Step 1 DIVIDE BOTH SIDES BY a TO MAKE THE LEADING COEFFICIENT 1.

    x26x+4=0This is the given equation.The leading coefficient,a,is 1,so we can skip this step.
  2. Step 2 ISOLATE THE VARIABLE TERMS ON ONE SIDE OF THE EQUATION AND THE CONSTANT TERM ON THE OTHER SIDE.

    x26x=4Subtract 4 from both sides.Now that x26x,the binomial,is isolated,we can complete the square.
  3. Step 3 (a) COMPLETE THE SQUARE. ADD THE SQUARE OF HALF THE COEFFICIENT OF x TO BOTH SIDES OF THE EQUATION.

    x26x+9=4+9We work with x26x=4. Half of 6 is 3 and (3)2=9.
  4. Step 3 (b) FACTOR THE RESULTING PERFECT SQUARE TRINOMIAL.

    (x3)2=5Factor  x26x+9  using  A22AB+B2=(AB)2.Simplify the right side:4+9=5.
  5. Step 4 USE THE SQUARE ROOT PROPERTY AND SOLVE FOR x.

    x3=5orx3=5Apply the square root property..3x=3+5or.x=35Add 3 to both sides in each equation.

    The solutions are 3±5 and the solution set is {3+5, 35}, or {3±5}.

Check Point 9

  • Solve by completing the square: x2+4x1=0.

Objective 9: Solve quadratic equations using the quadratic formula

Solving Quadratic Equations Using the Quadratic Formula

  1. Objective 9Solve quadratic equations using the quadratic formula.

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We can use the method of completing the square to derive a formula that can be used to solve all quadratic equations. The derivation given below also shows a particular quadratic equation, 3x22x4=0, to specifically illustrate each of the steps.

Deriving the Quadratic Formula

A table titled, deriving the quadratic formula with corresponding examples.
P.8-172 Full Alternative Text

The formula shown at the bottom of the left column is called the quadratic formula. A similar proof shows that the same formula can be used to solve quadratic equations if a, the coefficient of the x2-term, is negative.

The Quadratic Formula

The solutions of a quadratic equation in standard form ax2+bx+c=0, with a0, are given by the quadratic formula:

An image shows a mathematical expression x = Start fraction negative b plus or minus the square root of start expression b squared minus 4 a c end expression over 2 a end fraction.

To use the quadratic formula, rewrite the quadratic equation in standard form if necessary. Then determine the numerical values for a (the coefficient of the x2-term), b (the coefficient of the x-term), and c (the constant term). Substitute the values of a, b, and c into the quadratic formula and evaluate the expression. The ± sign indicates that there are two (not necessarily distinct) solutions of the equation.

Example 10 Solving a Quadratic Equation Using the Quadratic Formula

Solve using the quadratic formula: 2x26x+1=0.

Solution

The given equation is in standard form. Begin by identifying the values for a, b, and c.

2 x squared minus 6 x + 1 = 0. Three dialogue box appears below pointing at 2 reads, a = 2, Negative 6 reads, b = negative 6. + 1 reads, c = 1.

Substituting these values into the quadratic formula and simplifying gives the equation’s solutions.

x=b±b24ac2aUse the quadratic formula.=(6)±(6)24(2)(1)22Substitute the values for a,b, and c:a=2,b=6, and c=1.=6±3684(6)=6,(6)2=(6)(6)=36, and4(2)(1)=8.=6±284Complete the subtraction under the radical.=6±27428=47=4 7=27=2(3±7)4Factor out 2 from the numerator.=3±72Divide the numerator and denominator by 2.

The solution set is {3+72, 372} or {3±72}.

Check Point 10

  • Solve using the quadratic formula:

    2x2+2x1=0.
Objective 9: Solve quadratic equations using the quadratic formula

Solving Quadratic Equations Using the Quadratic Formula

  1. Objective 9Solve quadratic equations using the quadratic formula.

Watch Video

We can use the method of completing the square to derive a formula that can be used to solve all quadratic equations. The derivation given below also shows a particular quadratic equation, 3x22x4=0, to specifically illustrate each of the steps.

Deriving the Quadratic Formula

A table titled, deriving the quadratic formula with corresponding examples.
P.8-172 Full Alternative Text

The formula shown at the bottom of the left column is called the quadratic formula. A similar proof shows that the same formula can be used to solve quadratic equations if a, the coefficient of the x2-term, is negative.

The Quadratic Formula

The solutions of a quadratic equation in standard form ax2+bx+c=0, with a0, are given by the quadratic formula:

An image shows a mathematical expression x = Start fraction negative b plus or minus the square root of start expression b squared minus 4 a c end expression over 2 a end fraction.

To use the quadratic formula, rewrite the quadratic equation in standard form if necessary. Then determine the numerical values for a (the coefficient of the x2-term), b (the coefficient of the x-term), and c (the constant term). Substitute the values of a, b, and c into the quadratic formula and evaluate the expression. The ± sign indicates that there are two (not necessarily distinct) solutions of the equation.

Example 10 Solving a Quadratic Equation Using the Quadratic Formula

Solve using the quadratic formula: 2x26x+1=0.

Solution

The given equation is in standard form. Begin by identifying the values for a, b, and c.

2 x squared minus 6 x + 1 = 0. Three dialogue box appears below pointing at 2 reads, a = 2, Negative 6 reads, b = negative 6. + 1 reads, c = 1.

Substituting these values into the quadratic formula and simplifying gives the equation’s solutions.

x=b±b24ac2aUse the quadratic formula.=(6)±(6)24(2)(1)22Substitute the values for a,b, and c:a=2,b=6, and c=1.=6±3684(6)=6,(6)2=(6)(6)=36, and4(2)(1)=8.=6±284Complete the subtraction under the radical.=6±27428=47=4 7=27=2(3±7)4Factor out 2 from the numerator.=3±72Divide the numerator and denominator by 2.

The solution set is {3+72, 372} or {3±72}.

Check Point 10

  • Solve using the quadratic formula:

    2x2+2x1=0.
Objective 10: Use the discriminant to determine the number and type of solutions of quadratic equations

Quadratic Equations and the Discriminant

  1. Objective 10Use the discriminant to determine the number and type of solutions of quadratic equations.

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The quantity b24ac, which appears under the radical sign in the quadratic formula, is called the discriminant. Table P.4 shows how the discriminant of the quadratic equation ax2+bx+c=0 determines the number and type of solutions.

Table P.4 The Discriminant and the Kinds of Solutions to ax2+bx+c=0

Discriminant b24ac Kinds of Solutions to ax2+bx+c=0
b24ac>0 Two unequal real solutions: If a, b, and c are rational numbers and the discriminant is a perfect square, the solutions are rational. If the discriminant is not a perfect square, the solutions are irrational.
b24ac=0 One solution (a repeated solution) that is a real number: If a, b, and c are rational numbers, the repeated solution is also a rational number.
b24ac<0 No real solutions

Example 11 Using the Discriminant

Compute the discriminant of 4x28x+1=0. What does the discriminant indicate about the number and type of solutions?

Solution

Begin by identifying the values for a, b, and c.

4 x squared minus 8 x + 1 = 0. A dialogue box appears below pointing at 4 reads, a = 4, negative 8x reads, b = negative 8, + 1 reads, c = 1.

Substitute and compute the discriminant:

b24ac=(8)2441=6416=48.

The discriminant is 48. Because the discriminant is positive, the equation 4x28x+1=0 has two unequal real solutions.

Check Point 11

  • Compute the discriminant of 3x22x+5=0. What does the discriminant indicate about the number and type of solutions?

Objective 11: Determine the most efficient method to use when solving a quadratic equation

Determining Which Method to Use

  1. Objective 11Determine the most efficient method to use when solving a quadratic equation.

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All quadratic equations can be solved by the quadratic formula. However, if an equation is in the form u2=d, such as x2=5 or (2x+3)2=8, it is faster to use the square root property, taking the square root of both sides. If the equation is not in the form u2=d, write the quadratic equation in standard form (ax2+bx+c=0). Try to solve the equation by factoring. If ax2+bx+c cannot be factored, then solve the quadratic equation by the quadratic formula.

Because we used the method of completing the square to derive the quadratic formula, we no longer need it for solving quadratic equations. However, we will use completing the square later in the book to help graph other kinds of equations.

Table P.5 summarizes our observations about which technique to use when solving a quadratic equation.

Table P.5 Determining the Most Efficient Technique to Use When Solving a Quadratic Equation

A table titled, Determining the most efficient technique to use when solving a quadratic equation with corresponding examples.
Table P.5 Full Alternative Text
Objective 12: Solve radical equations

Radical Equations

  1. Objective 12Solve radical equations.

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A radical equation is an equation in which the variable occurs in a square root, cube root, or any higher root. An example of a radical equation is

x=9.

We solve x=9 by squaring both sides:

Left parenthesis radical x right parenthesis squared = 9 squared, x = 81.A note reads, squaring both sides eliminates the square roots.

The proposed solution, 81, can be checked in the original equation, x=9. Because 81=9, the solution is 81 and the solution set is {81}.

In general, we solve radical equations with square roots by squaring both sides of the equation. We solve radical equations with nth roots by raising both sides of the equation to the nth power. Unfortunately, if n is even, all the solutions of the equation raised to the even power may not be solutions of the original equation. Consider, for example, the equation

x=4.

If we square both sides, we obtain

x2=16.

Solving this equation using the square root property, we obtain

x=±16=±4.

The new equation x2=16 has two solutions, 4 and 4. By contrast, only 4 is a solution of the original equation, x=4. For this reason, when raising both sides of an equation to an even power, always check proposed solutions in the original equation.

Here is a general method for solving radical equations with nth roots:

Solving Radical Equations Containing nth Roots

  1. If necessary, arrange terms so that one radical is isolated on one side of the equation.

  2. Raise both sides of the equation to the nth power to eliminate the isolated nth root.

  3. Solve the resulting equation. If this equation still contains radicals, repeat steps 1 and 2.

  4. Check all proposed solutions in the original equation.

Extra solutions may be introduced when you raise both sides of a radical equation to an even power. Such solutions, which are not solutions of the given equation, are called extraneous solutions or extraneous roots.

Example 12 Solving a Radical Equation

Solve: 2x1+2=x.

Solution

  1. Step 1 ISOLATE A RADICAL ON ONE SIDE. We isolate the radical, 2x1, by subtracting 2 from both sides.

    2x1+2=xThis is the given equation.2x1=x2Subtract 2 from both sides.
  2. Step 2 RAISE BOTH SIDES TO THE nTH POWER. Because n, the index of the radical in the equation 2x1 = x2, is 2, we square both sides.

    (2x1)2=(x2)22x1=x24x+4Simplify. Use the formula(AB)2=A22AB+B2 on theright side.

  3. Step 3 SOLVE THE RESULTING EQUATION. Because of the x2-term, the resulting equation is a quadratic equation. We can obtain 0 on the left side by subtracting 2x and adding 1 on both sides.

    2x1=x24x+4The resulting equation is quadratic.0=x26x+5Write in standard from, subtracting 2xand adding 1 on both sides.0=(x1)(x5)Factor.x1=0 orx5=0Set each factor equal to 0.x=1 or5x=5Solve the resulting equations.
  4. Step 4 CHECK THE PROPOSED SOLUTIONS IN THE ORIGINAL EQUATION.

    Check 1:

    2x1+2=x,211+2=?1,1+2=?1,1+2=?1,3=1,false

    Check 5:

    2x1+2=x,251+2=?5,9+2=?5,3+2=?5,5=5,true

    Thus, 1 is an extraneous solution. The only solution is 5, and the solution set is {5}.

Check Point 12

  • Solve: x+3+3=x.

Objective 12: Solve radical equations

Radical Equations

  1. Objective 12Solve radical equations.

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A radical equation is an equation in which the variable occurs in a square root, cube root, or any higher root. An example of a radical equation is

x=9.

We solve x=9 by squaring both sides:

Left parenthesis radical x right parenthesis squared = 9 squared, x = 81.A note reads, squaring both sides eliminates the square roots.

The proposed solution, 81, can be checked in the original equation, x=9. Because 81=9, the solution is 81 and the solution set is {81}.

In general, we solve radical equations with square roots by squaring both sides of the equation. We solve radical equations with nth roots by raising both sides of the equation to the nth power. Unfortunately, if n is even, all the solutions of the equation raised to the even power may not be solutions of the original equation. Consider, for example, the equation

x=4.

If we square both sides, we obtain

x2=16.

Solving this equation using the square root property, we obtain

x=±16=±4.

The new equation x2=16 has two solutions, 4 and 4. By contrast, only 4 is a solution of the original equation, x=4. For this reason, when raising both sides of an equation to an even power, always check proposed solutions in the original equation.

Here is a general method for solving radical equations with nth roots:

Solving Radical Equations Containing nth Roots

  1. If necessary, arrange terms so that one radical is isolated on one side of the equation.

  2. Raise both sides of the equation to the nth power to eliminate the isolated nth root.

  3. Solve the resulting equation. If this equation still contains radicals, repeat steps 1 and 2.

  4. Check all proposed solutions in the original equation.

Extra solutions may be introduced when you raise both sides of a radical equation to an even power. Such solutions, which are not solutions of the given equation, are called extraneous solutions or extraneous roots.

Example 12 Solving a Radical Equation

Solve: 2x1+2=x.

Solution

  1. Step 1 ISOLATE A RADICAL ON ONE SIDE. We isolate the radical, 2x1, by subtracting 2 from both sides.

    2x1+2=xThis is the given equation.2x1=x2Subtract 2 from both sides.
  2. Step 2 RAISE BOTH SIDES TO THE nTH POWER. Because n, the index of the radical in the equation 2x1 = x2, is 2, we square both sides.

    (2x1)2=(x2)22x1=x24x+4Simplify. Use the formula(AB)2=A22AB+B2 on theright side.

  3. Step 3 SOLVE THE RESULTING EQUATION. Because of the x2-term, the resulting equation is a quadratic equation. We can obtain 0 on the left side by subtracting 2x and adding 1 on both sides.

    2x1=x24x+4The resulting equation is quadratic.0=x26x+5Write in standard from, subtracting 2xand adding 1 on both sides.0=(x1)(x5)Factor.x1=0 orx5=0Set each factor equal to 0.x=1 or5x=5Solve the resulting equations.
  4. Step 4 CHECK THE PROPOSED SOLUTIONS IN THE ORIGINAL EQUATION.

    Check 1:

    2x1+2=x,211+2=?1,1+2=?1,1+2=?1,3=1,false

    Check 5:

    2x1+2=x,251+2=?5,9+2=?5,3+2=?5,5=5,true

    Thus, 1 is an extraneous solution. The only solution is 5, and the solution set is {5}.

Check Point 12

  • Solve: x+3+3=x.

P.7: Concept and Vocabulary Check

P.7 Concept and Vocabulary Check

Fill in each blank so that the resulting statement is true.

  1. C1. An equation in the form ax+b=0, a0, such as 3x+17=0, is called a/an ______ equation in one variable.

  2. C2. Two or more equations that have the same solution set are called _________ equations.

  3. C3. The first step in solving 7+3(x2)=2x+10 is to ________________.

  4. C4. The fractions in the equation

    x4=2+x33

    can be eliminated by multiplying both sides by the _______________ of x4 and x33, which is _____.

  5. C5. We reject any proposed solution of a rational equation that causes a denominator to equal ____.

  6. C6. The restrictions on the variable in the rational equation

    1x22x+4=2x1x2+2x8

    are ______ and ________.

  7. C7.

    5x+4+3x+3=12x+9(x+4)(x+3)(x+4)(x+3)(5x+4+3x+3)=(x+4)(x+3)(12x+9(x+4)(x+3))

    The resulting equation cleared of fractions is ____________.

  8. C8. Solving a formula for a variable means rewriting the formula so that the variable is ________.

  9. C9. The first step in solving IR+Ir=E for I is to obtain a single occurrence of I by _________ I from the two terms on the left.

  10. C10. If c>0,|u|=c is equivalent to u= _______ or u= _______.

  11. C11. |3x1|=7 is equivalent to _________ or __________.

  12. C12. An equation that can be written in the standard form ax2+bx+c=0, a0, is called a/an _________ equation.

  13. C13. The zero-product principle states that if AB=0, then ________.

  14. C14. The square root property states that if u2=d, then u= _______.

  15. C15. If x2=7, then x= _______.

  16. C16. To solve x2+6x=7 by completing the square, add ____ to both sides of the equation.

  17. C17. The solutions of a quadratic equation in the standard form ax2+bx+c=0, a0, are given by the quadratic formula x= _______.

  18. C18. In order to solve 2x2+9x5=0 by the quadratic formula, we use a= _______, b= _______, and c= _______.

  19. C19. In order to solve x2=4x+1 by the quadratic formula, we use a= _______, b= _______, and c= _______.

  20. C20. x=(4)±(4)24(1)(2)2(1) simplifies to x= _______.

  21. C21. The discriminant of ax2+bx+c=0 is defined by _________.

  22. C22. If the discriminant of ax2+bx+c=0 is negative, the quadratic equation has _____ real solutions.

  23. C23. If the discriminant of ax2+bx+c=0 is positive, the quadratic equation has ______ real solutions.

  24. C24. The most efficient technique for solving (2x+7)2=25 is by using __________.

  25. C25. The most efficient technique for solving x2+5x10=0 is by using ____________.

  26. C26. The most efficient technique for solving x2 + 8x + 15 = 0 is by using __________________.

  27. C27. An equation in which the variable occurs in a square root, cube root, or any higher root is called a/an _______ equation.

  28. C28. Solutions of a squared equation that are not solutions of the original equation are called _________ solutions.

  29. C29. Consider the equation

    2x+1=x7.

    Squaring the left side and simplifying results in ______. Squaring the right side and simplifying results in ______.

P.7: Exercise Set

P.7 Exercise Set

Practice Exercises

In Exercises 116, solve and check each linear equation.

  1. 1. 7x5=72

  2. 2. 6x3=63

  3. 3. 11x(6x5)=40

  4. 4. 5x(2x10)=35

  5. 5. 2x7=6+x

  6. 6. 3x+5=2x+13

  7. 7. 7x+4=x+16

  8. 8. 13x+14=12x5

  9. 9. 3(x2)+7=2(x+5)

  10. 10. 2(x1)+3=x3(x+1)

  11. 11. x+36=38+x54

  12. 12. x+14=16+2x3

  13. 13. x4=2+x33

  14. 14. 5+x23=x+38

  15. 15. x+13=5x+27

  16. 16. 3x5x32=x+23

Exercises 1726 contain rational equations with variables in denominators. For each equation, a. Write the value or values of the variable that make a denominator zero. These are the restrictions on the variable. b. Keeping the restrictions in mind, solve the equation.

  1. 17. 1x1+5=11x1

  2. 18. 3x+47=4x+4

  3. 19. 8xx+1=48x+1

  4. 20. 2x2=xx22

  5. 21. 32x2+12=2x1

  6. 22. 3x+3=52x+6+1x2

  7. 23. 2x+11x1=2xx21

  8. 24. 4x+5+2x5=32x225

  9. 25. 1x45x+2=6x22x8

  10. 26. 1x32x+1=8x22x3

In Exercises 2742, solve each formula for the specified variable. Do you recognize the formula? If so, what does it describe?

  1. 27. I=Prt for P

  2. 28. C=2πr for r

  3. 29. T=D+pm for p

  4. 30. P=C+MC for M

  5. 31. A=12h(a+b) for a

  6. 32. A=12h(a+b) for b

  7. 33. S=P+Prt for r

  8. 34. S=P+Prt for t

  9. 35. B=FSV for S

  10. 36. S=C1r for r

  11. 37. IR+Ir=E for I

  12. 38. A=2lw+2lh+2wh for h

  13. 39. 1p+1q=1f for f

  14. 40. 1R=1R1+1R2 for R1

  15. 41. f=f1f2f1+f2 for f1

  16. 42. f=f1f2f1+f2 for f2

In Exercises 4354, solve each absolute value equation or indicate the equation has no solution.

  1. 43. |x2|=7

  2. 44. |x+1|=5

  3. 45. |2x1|=5

  4. 46. |2x3|=11

  5. 47. 2|3x2|=14

  6. 48. 3|2x1|=21

  7. 49. 2|452 x|+6=18

  8. 50. 4|134 x|+7=10

  9. 51. |x+1|+5=3

  10. 52. |x+1|+6=2

  11. 53. |2x1|+3=3

  12. 54. |3x2|+4=4

In Exercises 5560, solve each quadratic equation by factoring.

  1. 55. x23x10=0

  2. 56. x213x+36=0

  3. 57. x2=8x15

  4. 58. x2=11x10

  5. 59. 5x2=20x

  6. 60. 3x2=12x

In Exercises 6166, solve each quadratic equation by the square root property.

  1. 61. 3x2=27

  2. 62. 5x2=45

  3. 63. 5x2+1=51

  4. 64. 3x21=47

  5. 65. 3(x4)2=15

  6. 66. 3(x+4)2=21

In Exercises 6774, solve each quadratic equation by completing the square.

  1. 67. x2+6x=7

  2. 68. x2+6x=8

  3. 69. x22x=2

  4. 70. x2+4x=12

  5. 71. x26x11=0

  6. 72. x22x5=0

  7. 73. x2+4x+1=0

  8. 74. x2+6x5=0

In Exercises 7582, solve each quadratic equation using the quadratic formula.

  1. 75. x2+8x+15=0

  2. 76. x2+8x+12=0

  3. 77. x2+5x+3=0

  4. 78. x2+5x+2=0

  5. 79. 3x23x4=0

  6. 80. 5x2+x2=0

  7. 81. 4x2=2x+7

  8. 82. 3x2=6x1

Compute the discriminant of each equation in Exercises 8390. What does the discriminant indicate about the number and type of solutions?

  1. 83. x24x5=0

  2. 84. 4x22x+3=0

  3. 85. 2x211x+3=0

  4. 86. 2x2+11x6=0

  5. 87. x2=2x1

  6. 88. 3x2=2x1

  7. 89. x23x7=0

  8. 90. 3x2+4x2=0

In Exercises 91114, solve each quadratic equation by the method of your choice.

  1. 91. 2x2x=1

  2. 92. 3x24x=4

  3. 93. 5x2+2=11x

  4. 94. 5x2=613x

  5. 95. 3x2=60

  6. 96. 2x2=250

  7. 97. x22x=1

  8. 98. 2x2+3x=1

  9. 99. (2x+3)(x+4)=1

  10. 100. (2x5)(x+1)=2

  11. 101. (3x4)2=16

  12. 102. (2x+7)2=25

  13. 103. 3x212x+12=0

  14. 104. 96x+x2=0

  15. 105. 4x216=0

  16. 106. 3x227=0

  17. 107. x2=4x2

  18. 108. x2=6x7

  19. 109. 2x27x=0

  20. 110. 2x2+5x=3

  21. 111. 1x+1x+2=13

  22. 112. 1x+1x+3=14

  23. 113. 2xx3+6x+3=28x29

  24. 114. 3x3+5x4=x220x27x+12

In Exercises 115124, solve each radical equation. Check all proposed solutions.

  1. 115. 3x+18=x

  2. 116. 208x=x

  3. 117. x+3=x3

  4. 118. x+10=x2

  5. 119. 2x+13=x+7

  6. 120. 6x+1=x1

  7. 121. x2x+5=5

  8. 122. xx+11=1

  9. 123. 2x+198=x

  10. 124. 2x+156=x

Practice PLUS

In Exercises 125134, solve each equation.

  1. 125. 25[2+5x3(x+2)]=3(2x5)[5(x1)3x+3]

  2. 126. 45[42x4(x+7)]=4(1+3x)[43(x+2)2(2x5)]

  3. 127. 77x=(3x+2)(x1)

  4. 128. 10x1=(2x+1)2

  5. 129. |x2+2x36|=12

  6. 130. |x2+6x+1|=8

  7. 131. 1x23x+2=1x+2+5x24

  8. 132. x1x2+xx3=1x25x+6

  9. 133. x+8x4=2

  10. 134. x+5x3=2

In Exercises 135136, list all numbers that must be excluded from the domain of each rational expression.

  1. 135. 32x2+4x9

  2. 136. 72x28x+5

Application Exercises

Grade Inflation. The bar graph shows the percentage of U.S. college freshmen with an average grade of A in high school.

A bar graph depicts the percentage of U.S. college freshmen with an average grade of A left parenthesis A negative to A positive right parenthesis in high school.

Source: Higher Education Research Institute

The data displayed by the bar graph can be described by the mathematical model

p=4x5+25,

where x is the number of years after 1980 and p is the percentage of U.S. college freshmen who had an average grade of A in high school. Use this information to solve Exercises 137138.

  1. 137.

    1. According to the formula, in 2010, what percentage of U.S. college freshmen had an average grade of A in high school? Does this underestimate or overestimate the percent displayed by the bar graph? By how much?

    2. If trends shown by the formula continue, project when 57% of U.S. college freshmen will have had an average grade of A in high school.

  2. 138.

    1. According to the formula, in 2000, what percentage of U.S. college freshmen had an average grade of A in high school? Does this underestimate or overestimate the percent displayed by the bar graph? By how much?

    2. If trends shown by the formula continue, project when 65% of U.S. college freshmen will have had an average grade of A in high school.

  3. 139. A company wants to increase the 10% peroxide content of its product by adding pure peroxide (100% peroxide). If x liters of pure peroxide are added to 500 liters of its 10% solution, the concentration, C, of the new mixture is given by

    C=x+0.1(500)x+500.

    How many liters of pure peroxide should be added to produce a new product that is 28% peroxide?

  4. 140. Suppose that x liters of pure acid are added to 200 liters of a 35% acid solution.

    1. Write a formula that gives the concentration, C, of the new mixture. (Hint: See Exercise 139.)

    2. How many liters of pure acid should be added to produce a new mixture that is 74% acid?

A driver’s age has something to do with his or her chance of getting into a fatal car crash. The bar graph shows the number of fatal vehicle crashes per 100 million miles driven for drivers of various age groups. For example, 25-year-old drivers are involved in 4.1 fatal crashes per 100 million miles driven. Thus, when a group of 25-year-old Americans have driven a total of 100 million miles, approximately 4 have been in accidents in which someone died.

A vertical bar graph titled, Age of United States drivers and fatal crashes.

Source: Insurance Institute for Highway Safety

The number of fatal vehicle crashes per 100 million miles, N, for drivers of age x can be modeled by the formula

N=0.013x21.19x+28.24.

Use the formula to solve Exercises 141142.

  1. 141. What age groups are expected to be involved in 3 fatal crashes per 100 million miles driven? How well does the formula model the trend in the actual data shown in the bar graph?

  2. 142. What age groups are expected to be involved in 10 fatal crashes per 100 million miles driven? How well does the formula model the trend in the actual data shown in the bar graph?

The graphs show the percentage of jobs in the U.S. labor force held by men and by women from 1970 through 2015. Exercises 143144 are based on the data displayed by the graphs.

The line graph titled, Percentage of U.S Jobs held by Men and Women.

Source: Bureau of Labor Statistics

  1. 143. The formula

    p=1.6t+38

    models the percentage of jobs in the U.S. labor force, p, held by women t years after 1970.

    1. Use the appropriate graph to estimate the percentage of jobs in the U.S. labor force held by women in 2010. Give your estimation to the nearest percent.

    2. Use the mathematical model to determine the percentage of jobs in the U.S. labor force held by women in 2010. Round to the nearest tenth of a percent.

    3. According to the formula, when will 51% of jobs in the U.S. labor force be held by women? Round to the nearest year.

  2. 144. The formula

    p=1.6t+62

    models the percentage of jobs in the U.S. labor force, p, held by men t years after 1970.

    1. Use the appropriate graph to estimate the percentage of jobs in the U.S. labor force held by men in 2010. Give your estimation to the nearest percent.

    2. Use the mathematical model to determine the percentage of jobs in the U.S. labor force held by men in 2010. Round to the nearest tenth of a percent.

    3. According to the formula, when will 49% of jobs in the U.S. labor force be held by men? Round to the nearest year.

Explaining the Concepts

  1. 145. What is a linear equation in one variable? Give an example of this type of equation.

  2. 146. Explain how to determine the restrictions on the variable for the equation

    3x+5+4x2=7x2+3x6.
  3. 147. What does it mean to solve a formula for a variable?

  4. 148. Explain how to solve an equation involving absolute value.

  5. 149. Why does the procedure that you explained in Exercise 148 not apply to the equation |x2|=3? What is the solution set for this equation?

  6. 150. What is a quadratic equation?

  7. 151. Explain how to solve x2+6x+8=0 using factoring and the zero-product principle.

  8. 152. Explain how to solve x2+6x+8=0 by completing the square.

  9. 153. Explain how to solve x2+6x+8=0 using the quadratic formula.

  10. 154. How is the quadratic formula derived?

  11. 155. What is the discriminant and what information does it provide about a quadratic equation?

  12. 156. If you are given a quadratic equation, how do you determine which method to use to solve it?

  13. 157. In solving 2x1+2=x, why is it a good idea to isolate the radical term? What if we don’t do this and simply square each side? Describe what happens.

  14. 158. What is an extraneous solution to a radical equation?

Critical Thinking Exercises

Make Sense? In Exercises 159162, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 159. The model p=0.8x+25 describes the percentage of college freshmen with an A average in high school, p, x years after 1980, so I have to solve a linear equation to determine the percentage of college freshmen with an A average in high school in 2020.

  2. 160. Although I can solve 3x+15=14 by first subtracting 15 from both sides, I find it easier to begin by multiplying both sides by 20, the least common denominator.

  3. 161. Because I want to solve 25x2169=0 fairly quickly, I’ll use the quadratic formula.

  4. 162. When checking a radical equation’s proposed solution, I can substitute into the original equation or any equation that is part of the solution process.

In Exercises 163166, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 163. The equation (2x3)2=25 is equivalent to 2x3=5.

  2. 164. Every quadratic equation has two distinct numbers in its solution set.

  3. 165. The equations 3y1=11 and 3y7=5 are equivalent.

  4. 166. The equation ax2+c=0, a0, cannot be solved by the quadratic formula.

  5. 167. Find b such that 7x+4b+13=x will have a solution set given by {6}.

  6. 168. Write a quadratic equation in standard form whose solution set is {3, 5}.

  7. 169. Solve for C:V=CCSL N.

  8. 170. Solve for t:s=16t2+v0t.

Preview Exercises

Exercises 171173 will help you prepare for the material covered in the next section.

  1. 171. Jane’s salary exceeds Jim’s by $150 per week. If x represents Jim’s weekly salary, write an algebraic expression that models Jane’s weekly salary.

  2. 172. A convenience store sells a refillable stainless-steel travel mug for $20. Customers who buy the mug can refill it with coffee for just $0.99 on each visit. Write an algebraic expression that models the total cost of the mug and x refills.

  3. 173. If the width of a rectangle is represented by x and the length is represented by x+200, write a simplified algebraic expression that models the rectangle’s perimeter.

P.7: Exercise Set

P.7 Exercise Set

Practice Exercises

In Exercises 116, solve and check each linear equation.

  1. 1. 7x5=72

  2. 2. 6x3=63

  3. 3. 11x(6x5)=40

  4. 4. 5x(2x10)=35

  5. 5. 2x7=6+x

  6. 6. 3x+5=2x+13

  7. 7. 7x+4=x+16

  8. 8. 13x+14=12x5

  9. 9. 3(x2)+7=2(x+5)

  10. 10. 2(x1)+3=x3(x+1)

  11. 11. x+36=38+x54

  12. 12. x+14=16+2x3

  13. 13. x4=2+x33

  14. 14. 5+x23=x+38

  15. 15. x+13=5x+27

  16. 16. 3x5x32=x+23

Exercises 1726 contain rational equations with variables in denominators. For each equation, a. Write the value or values of the variable that make a denominator zero. These are the restrictions on the variable. b. Keeping the restrictions in mind, solve the equation.

  1. 17. 1x1+5=11x1

  2. 18. 3x+47=4x+4

  3. 19. 8xx+1=48x+1

  4. 20. 2x2=xx22

  5. 21. 32x2+12=2x1

  6. 22. 3x+3=52x+6+1x2

  7. 23. 2x+11x1=2xx21

  8. 24. 4x+5+2x5=32x225

  9. 25. 1x45x+2=6x22x8

  10. 26. 1x32x+1=8x22x3

In Exercises 2742, solve each formula for the specified variable. Do you recognize the formula? If so, what does it describe?

  1. 27. I=Prt for P

  2. 28. C=2πr for r

  3. 29. T=D+pm for p

  4. 30. P=C+MC for M

  5. 31. A=12h(a+b) for a

  6. 32. A=12h(a+b) for b

  7. 33. S=P+Prt for r

  8. 34. S=P+Prt for t

  9. 35. B=FSV for S

  10. 36. S=C1r for r

  11. 37. IR+Ir=E for I

  12. 38. A=2lw+2lh+2wh for h

  13. 39. 1p+1q=1f for f

  14. 40. 1R=1R1+1R2 for R1

  15. 41. f=f1f2f1+f2 for f1

  16. 42. f=f1f2f1+f2 for f2

In Exercises 4354, solve each absolute value equation or indicate the equation has no solution.

  1. 43. |x2|=7

  2. 44. |x+1|=5

  3. 45. |2x1|=5

  4. 46. |2x3|=11

  5. 47. 2|3x2|=14

  6. 48. 3|2x1|=21

  7. 49. 2|452 x|+6=18

  8. 50. 4|134 x|+7=10

  9. 51. |x+1|+5=3

  10. 52. |x+1|+6=2

  11. 53. |2x1|+3=3

  12. 54. |3x2|+4=4

In Exercises 5560, solve each quadratic equation by factoring.

  1. 55. x23x10=0

  2. 56. x213x+36=0

  3. 57. x2=8x15

  4. 58. x2=11x10

  5. 59. 5x2=20x

  6. 60. 3x2=12x

In Exercises 6166, solve each quadratic equation by the square root property.

  1. 61. 3x2=27

  2. 62. 5x2=45

  3. 63. 5x2+1=51

  4. 64. 3x21=47

  5. 65. 3(x4)2=15

  6. 66. 3(x+4)2=21

In Exercises 6774, solve each quadratic equation by completing the square.

  1. 67. x2+6x=7

  2. 68. x2+6x=8

  3. 69. x22x=2

  4. 70. x2+4x=12

  5. 71. x26x11=0

  6. 72. x22x5=0

  7. 73. x2+4x+1=0

  8. 74. x2+6x5=0

In Exercises 7582, solve each quadratic equation using the quadratic formula.

  1. 75. x2+8x+15=0

  2. 76. x2+8x+12=0

  3. 77. x2+5x+3=0

  4. 78. x2+5x+2=0

  5. 79. 3x23x4=0

  6. 80. 5x2+x2=0

  7. 81. 4x2=2x+7

  8. 82. 3x2=6x1

Compute the discriminant of each equation in Exercises 8390. What does the discriminant indicate about the number and type of solutions?

  1. 83. x24x5=0

  2. 84. 4x22x+3=0

  3. 85. 2x211x+3=0

  4. 86. 2x2+11x6=0

  5. 87. x2=2x1

  6. 88. 3x2=2x1

  7. 89. x23x7=0

  8. 90. 3x2+4x2=0

In Exercises 91114, solve each quadratic equation by the method of your choice.

  1. 91. 2x2x=1

  2. 92. 3x24x=4

  3. 93. 5x2+2=11x

  4. 94. 5x2=613x

  5. 95. 3x2=60

  6. 96. 2x2=250

  7. 97. x22x=1

  8. 98. 2x2+3x=1

  9. 99. (2x+3)(x+4)=1

  10. 100. (2x5)(x+1)=2

  11. 101. (3x4)2=16

  12. 102. (2x+7)2=25

  13. 103. 3x212x+12=0

  14. 104. 96x+x2=0

  15. 105. 4x216=0

  16. 106. 3x227=0

  17. 107. x2=4x2

  18. 108. x2=6x7

  19. 109. 2x27x=0

  20. 110. 2x2+5x=3

  21. 111. 1x+1x+2=13

  22. 112. 1x+1x+3=14

  23. 113. 2xx3+6x+3=28x29

  24. 114. 3x3+5x4=x220x27x+12

In Exercises 115124, solve each radical equation. Check all proposed solutions.

  1. 115. 3x+18=x

  2. 116. 208x=x

  3. 117. x+3=x3

  4. 118. x+10=x2

  5. 119. 2x+13=x+7

  6. 120. 6x+1=x1

  7. 121. x2x+5=5

  8. 122. xx+11=1

  9. 123. 2x+198=x

  10. 124. 2x+156=x

Practice PLUS

In Exercises 125134, solve each equation.

  1. 125. 25[2+5x3(x+2)]=3(2x5)[5(x1)3x+3]

  2. 126. 45[42x4(x+7)]=4(1+3x)[43(x+2)2(2x5)]

  3. 127. 77x=(3x+2)(x1)

  4. 128. 10x1=(2x+1)2

  5. 129. |x2+2x36|=12

  6. 130. |x2+6x+1|=8

  7. 131. 1x23x+2=1x+2+5x24

  8. 132. x1x2+xx3=1x25x+6

  9. 133. x+8x4=2

  10. 134. x+5x3=2

In Exercises 135136, list all numbers that must be excluded from the domain of each rational expression.

  1. 135. 32x2+4x9

  2. 136. 72x28x+5

Application Exercises

Grade Inflation. The bar graph shows the percentage of U.S. college freshmen with an average grade of A in high school.

A bar graph depicts the percentage of U.S. college freshmen with an average grade of A left parenthesis A negative to A positive right parenthesis in high school.

Source: Higher Education Research Institute

The data displayed by the bar graph can be described by the mathematical model

p=4x5+25,

where x is the number of years after 1980 and p is the percentage of U.S. college freshmen who had an average grade of A in high school. Use this information to solve Exercises 137138.

  1. 137.

    1. According to the formula, in 2010, what percentage of U.S. college freshmen had an average grade of A in high school? Does this underestimate or overestimate the percent displayed by the bar graph? By how much?

    2. If trends shown by the formula continue, project when 57% of U.S. college freshmen will have had an average grade of A in high school.

  2. 138.

    1. According to the formula, in 2000, what percentage of U.S. college freshmen had an average grade of A in high school? Does this underestimate or overestimate the percent displayed by the bar graph? By how much?

    2. If trends shown by the formula continue, project when 65% of U.S. college freshmen will have had an average grade of A in high school.

  3. 139. A company wants to increase the 10% peroxide content of its product by adding pure peroxide (100% peroxide). If x liters of pure peroxide are added to 500 liters of its 10% solution, the concentration, C, of the new mixture is given by

    C=x+0.1(500)x+500.

    How many liters of pure peroxide should be added to produce a new product that is 28% peroxide?

  4. 140. Suppose that x liters of pure acid are added to 200 liters of a 35% acid solution.

    1. Write a formula that gives the concentration, C, of the new mixture. (Hint: See Exercise 139.)

    2. How many liters of pure acid should be added to produce a new mixture that is 74% acid?

A driver’s age has something to do with his or her chance of getting into a fatal car crash. The bar graph shows the number of fatal vehicle crashes per 100 million miles driven for drivers of various age groups. For example, 25-year-old drivers are involved in 4.1 fatal crashes per 100 million miles driven. Thus, when a group of 25-year-old Americans have driven a total of 100 million miles, approximately 4 have been in accidents in which someone died.

A vertical bar graph titled, Age of United States drivers and fatal crashes.

Source: Insurance Institute for Highway Safety

The number of fatal vehicle crashes per 100 million miles, N, for drivers of age x can be modeled by the formula

N=0.013x21.19x+28.24.

Use the formula to solve Exercises 141142.

  1. 141. What age groups are expected to be involved in 3 fatal crashes per 100 million miles driven? How well does the formula model the trend in the actual data shown in the bar graph?

  2. 142. What age groups are expected to be involved in 10 fatal crashes per 100 million miles driven? How well does the formula model the trend in the actual data shown in the bar graph?

The graphs show the percentage of jobs in the U.S. labor force held by men and by women from 1970 through 2015. Exercises 143144 are based on the data displayed by the graphs.

The line graph titled, Percentage of U.S Jobs held by Men and Women.

Source: Bureau of Labor Statistics

  1. 143. The formula

    p=1.6t+38

    models the percentage of jobs in the U.S. labor force, p, held by women t years after 1970.

    1. Use the appropriate graph to estimate the percentage of jobs in the U.S. labor force held by women in 2010. Give your estimation to the nearest percent.

    2. Use the mathematical model to determine the percentage of jobs in the U.S. labor force held by women in 2010. Round to the nearest tenth of a percent.

    3. According to the formula, when will 51% of jobs in the U.S. labor force be held by women? Round to the nearest year.

  2. 144. The formula

    p=1.6t+62

    models the percentage of jobs in the U.S. labor force, p, held by men t years after 1970.

    1. Use the appropriate graph to estimate the percentage of jobs in the U.S. labor force held by men in 2010. Give your estimation to the nearest percent.

    2. Use the mathematical model to determine the percentage of jobs in the U.S. labor force held by men in 2010. Round to the nearest tenth of a percent.

    3. According to the formula, when will 49% of jobs in the U.S. labor force be held by men? Round to the nearest year.

Explaining the Concepts

  1. 145. What is a linear equation in one variable? Give an example of this type of equation.

  2. 146. Explain how to determine the restrictions on the variable for the equation

    3x+5+4x2=7x2+3x6.
  3. 147. What does it mean to solve a formula for a variable?

  4. 148. Explain how to solve an equation involving absolute value.

  5. 149. Why does the procedure that you explained in Exercise 148 not apply to the equation |x2|=3? What is the solution set for this equation?

  6. 150. What is a quadratic equation?

  7. 151. Explain how to solve x2+6x+8=0 using factoring and the zero-product principle.

  8. 152. Explain how to solve x2+6x+8=0 by completing the square.

  9. 153. Explain how to solve x2+6x+8=0 using the quadratic formula.

  10. 154. How is the quadratic formula derived?

  11. 155. What is the discriminant and what information does it provide about a quadratic equation?

  12. 156. If you are given a quadratic equation, how do you determine which method to use to solve it?

  13. 157. In solving 2x1+2=x, why is it a good idea to isolate the radical term? What if we don’t do this and simply square each side? Describe what happens.

  14. 158. What is an extraneous solution to a radical equation?

Critical Thinking Exercises

Make Sense? In Exercises 159162, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 159. The model p=0.8x+25 describes the percentage of college freshmen with an A average in high school, p, x years after 1980, so I have to solve a linear equation to determine the percentage of college freshmen with an A average in high school in 2020.

  2. 160. Although I can solve 3x+15=14 by first subtracting 15 from both sides, I find it easier to begin by multiplying both sides by 20, the least common denominator.

  3. 161. Because I want to solve 25x2169=0 fairly quickly, I’ll use the quadratic formula.

  4. 162. When checking a radical equation’s proposed solution, I can substitute into the original equation or any equation that is part of the solution process.

In Exercises 163166, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 163. The equation (2x3)2=25 is equivalent to 2x3=5.

  2. 164. Every quadratic equation has two distinct numbers in its solution set.

  3. 165. The equations 3y1=11 and 3y7=5 are equivalent.

  4. 166. The equation ax2+c=0, a0, cannot be solved by the quadratic formula.

  5. 167. Find b such that 7x+4b+13=x will have a solution set given by {6}.

  6. 168. Write a quadratic equation in standard form whose solution set is {3, 5}.

  7. 169. Solve for C:V=CCSL N.

  8. 170. Solve for t:s=16t2+v0t.

Preview Exercises

Exercises 171173 will help you prepare for the material covered in the next section.

  1. 171. Jane’s salary exceeds Jim’s by $150 per week. If x represents Jim’s weekly salary, write an algebraic expression that models Jane’s weekly salary.

  2. 172. A convenience store sells a refillable stainless-steel travel mug for $20. Customers who buy the mug can refill it with coffee for just $0.99 on each visit. Write an algebraic expression that models the total cost of the mug and x refills.

  3. 173. If the width of a rectangle is represented by x and the length is represented by x+200, write a simplified algebraic expression that models the rectangle’s perimeter.

P.7: Exercise Set

P.7 Exercise Set

Practice Exercises

In Exercises 116, solve and check each linear equation.

  1. 1. 7x5=72

  2. 2. 6x3=63

  3. 3. 11x(6x5)=40

  4. 4. 5x(2x10)=35

  5. 5. 2x7=6+x

  6. 6. 3x+5=2x+13

  7. 7. 7x+4=x+16

  8. 8. 13x+14=12x5

  9. 9. 3(x2)+7=2(x+5)

  10. 10. 2(x1)+3=x3(x+1)

  11. 11. x+36=38+x54

  12. 12. x+14=16+2x3

  13. 13. x4=2+x33

  14. 14. 5+x23=x+38

  15. 15. x+13=5x+27

  16. 16. 3x5x32=x+23

Exercises 1726 contain rational equations with variables in denominators. For each equation, a. Write the value or values of the variable that make a denominator zero. These are the restrictions on the variable. b. Keeping the restrictions in mind, solve the equation.

  1. 17. 1x1+5=11x1

  2. 18. 3x+47=4x+4

  3. 19. 8xx+1=48x+1

  4. 20. 2x2=xx22

  5. 21. 32x2+12=2x1

  6. 22. 3x+3=52x+6+1x2

  7. 23. 2x+11x1=2xx21

  8. 24. 4x+5+2x5=32x225

  9. 25. 1x45x+2=6x22x8

  10. 26. 1x32x+1=8x22x3

In Exercises 2742, solve each formula for the specified variable. Do you recognize the formula? If so, what does it describe?

  1. 27. I=Prt for P

  2. 28. C=2πr for r

  3. 29. T=D+pm for p

  4. 30. P=C+MC for M

  5. 31. A=12h(a+b) for a

  6. 32. A=12h(a+b) for b

  7. 33. S=P+Prt for r

  8. 34. S=P+Prt for t

  9. 35. B=FSV for S

  10. 36. S=C1r for r

  11. 37. IR+Ir=E for I

  12. 38. A=2lw+2lh+2wh for h

  13. 39. 1p+1q=1f for f

  14. 40. 1R=1R1+1R2 for R1

  15. 41. f=f1f2f1+f2 for f1

  16. 42. f=f1f2f1+f2 for f2

In Exercises 4354, solve each absolute value equation or indicate the equation has no solution.

  1. 43. |x2|=7

  2. 44. |x+1|=5

  3. 45. |2x1|=5

  4. 46. |2x3|=11

  5. 47. 2|3x2|=14

  6. 48. 3|2x1|=21

  7. 49. 2|452 x|+6=18

  8. 50. 4|134 x|+7=10

  9. 51. |x+1|+5=3

  10. 52. |x+1|+6=2

  11. 53. |2x1|+3=3

  12. 54. |3x2|+4=4

In Exercises 5560, solve each quadratic equation by factoring.

  1. 55. x23x10=0

  2. 56. x213x+36=0

  3. 57. x2=8x15

  4. 58. x2=11x10

  5. 59. 5x2=20x

  6. 60. 3x2=12x

In Exercises 6166, solve each quadratic equation by the square root property.

  1. 61. 3x2=27

  2. 62. 5x2=45

  3. 63. 5x2+1=51

  4. 64. 3x21=47

  5. 65. 3(x4)2=15

  6. 66. 3(x+4)2=21

In Exercises 6774, solve each quadratic equation by completing the square.

  1. 67. x2+6x=7

  2. 68. x2+6x=8

  3. 69. x22x=2

  4. 70. x2+4x=12

  5. 71. x26x11=0

  6. 72. x22x5=0

  7. 73. x2+4x+1=0

  8. 74. x2+6x5=0

In Exercises 7582, solve each quadratic equation using the quadratic formula.

  1. 75. x2+8x+15=0

  2. 76. x2+8x+12=0

  3. 77. x2+5x+3=0

  4. 78. x2+5x+2=0

  5. 79. 3x23x4=0

  6. 80. 5x2+x2=0

  7. 81. 4x2=2x+7

  8. 82. 3x2=6x1

Compute the discriminant of each equation in Exercises 8390. What does the discriminant indicate about the number and type of solutions?

  1. 83. x24x5=0

  2. 84. 4x22x+3=0

  3. 85. 2x211x+3=0

  4. 86. 2x2+11x6=0

  5. 87. x2=2x1

  6. 88. 3x2=2x1

  7. 89. x23x7=0

  8. 90. 3x2+4x2=0

In Exercises 91114, solve each quadratic equation by the method of your choice.

  1. 91. 2x2x=1

  2. 92. 3x24x=4

  3. 93. 5x2+2=11x

  4. 94. 5x2=613x

  5. 95. 3x2=60

  6. 96. 2x2=250

  7. 97. x22x=1

  8. 98. 2x2+3x=1

  9. 99. (2x+3)(x+4)=1

  10. 100. (2x5)(x+1)=2

  11. 101. (3x4)2=16

  12. 102. (2x+7)2=25

  13. 103. 3x212x+12=0

  14. 104. 96x+x2=0

  15. 105. 4x216=0

  16. 106. 3x227=0

  17. 107. x2=4x2

  18. 108. x2=6x7

  19. 109. 2x27x=0

  20. 110. 2x2+5x=3

  21. 111. 1x+1x+2=13

  22. 112. 1x+1x+3=14

  23. 113. 2xx3+6x+3=28x29

  24. 114. 3x3+5x4=x220x27x+12

In Exercises 115124, solve each radical equation. Check all proposed solutions.

  1. 115. 3x+18=x

  2. 116. 208x=x

  3. 117. x+3=x3

  4. 118. x+10=x2

  5. 119. 2x+13=x+7

  6. 120. 6x+1=x1

  7. 121. x2x+5=5

  8. 122. xx+11=1

  9. 123. 2x+198=x

  10. 124. 2x+156=x

Practice PLUS

In Exercises 125134, solve each equation.

  1. 125. 25[2+5x3(x+2)]=3(2x5)[5(x1)3x+3]

  2. 126. 45[42x4(x+7)]=4(1+3x)[43(x+2)2(2x5)]

  3. 127. 77x=(3x+2)(x1)

  4. 128. 10x1=(2x+1)2

  5. 129. |x2+2x36|=12

  6. 130. |x2+6x+1|=8

  7. 131. 1x23x+2=1x+2+5x24

  8. 132. x1x2+xx3=1x25x+6

  9. 133. x+8x4=2

  10. 134. x+5x3=2

In Exercises 135136, list all numbers that must be excluded from the domain of each rational expression.

  1. 135. 32x2+4x9

  2. 136. 72x28x+5

Application Exercises

Grade Inflation. The bar graph shows the percentage of U.S. college freshmen with an average grade of A in high school.

A bar graph depicts the percentage of U.S. college freshmen with an average grade of A left parenthesis A negative to A positive right parenthesis in high school.

Source: Higher Education Research Institute

The data displayed by the bar graph can be described by the mathematical model

p=4x5+25,

where x is the number of years after 1980 and p is the percentage of U.S. college freshmen who had an average grade of A in high school. Use this information to solve Exercises 137138.

  1. 137.

    1. According to the formula, in 2010, what percentage of U.S. college freshmen had an average grade of A in high school? Does this underestimate or overestimate the percent displayed by the bar graph? By how much?

    2. If trends shown by the formula continue, project when 57% of U.S. college freshmen will have had an average grade of A in high school.

  2. 138.

    1. According to the formula, in 2000, what percentage of U.S. college freshmen had an average grade of A in high school? Does this underestimate or overestimate the percent displayed by the bar graph? By how much?

    2. If trends shown by the formula continue, project when 65% of U.S. college freshmen will have had an average grade of A in high school.

  3. 139. A company wants to increase the 10% peroxide content of its product by adding pure peroxide (100% peroxide). If x liters of pure peroxide are added to 500 liters of its 10% solution, the concentration, C, of the new mixture is given by

    C=x+0.1(500)x+500.

    How many liters of pure peroxide should be added to produce a new product that is 28% peroxide?

  4. 140. Suppose that x liters of pure acid are added to 200 liters of a 35% acid solution.

    1. Write a formula that gives the concentration, C, of the new mixture. (Hint: See Exercise 139.)

    2. How many liters of pure acid should be added to produce a new mixture that is 74% acid?

A driver’s age has something to do with his or her chance of getting into a fatal car crash. The bar graph shows the number of fatal vehicle crashes per 100 million miles driven for drivers of various age groups. For example, 25-year-old drivers are involved in 4.1 fatal crashes per 100 million miles driven. Thus, when a group of 25-year-old Americans have driven a total of 100 million miles, approximately 4 have been in accidents in which someone died.

A vertical bar graph titled, Age of United States drivers and fatal crashes.

Source: Insurance Institute for Highway Safety

The number of fatal vehicle crashes per 100 million miles, N, for drivers of age x can be modeled by the formula

N=0.013x21.19x+28.24.

Use the formula to solve Exercises 141142.

  1. 141. What age groups are expected to be involved in 3 fatal crashes per 100 million miles driven? How well does the formula model the trend in the actual data shown in the bar graph?

  2. 142. What age groups are expected to be involved in 10 fatal crashes per 100 million miles driven? How well does the formula model the trend in the actual data shown in the bar graph?

The graphs show the percentage of jobs in the U.S. labor force held by men and by women from 1970 through 2015. Exercises 143144 are based on the data displayed by the graphs.

The line graph titled, Percentage of U.S Jobs held by Men and Women.

Source: Bureau of Labor Statistics

  1. 143. The formula

    p=1.6t+38

    models the percentage of jobs in the U.S. labor force, p, held by women t years after 1970.

    1. Use the appropriate graph to estimate the percentage of jobs in the U.S. labor force held by women in 2010. Give your estimation to the nearest percent.

    2. Use the mathematical model to determine the percentage of jobs in the U.S. labor force held by women in 2010. Round to the nearest tenth of a percent.

    3. According to the formula, when will 51% of jobs in the U.S. labor force be held by women? Round to the nearest year.

  2. 144. The formula

    p=1.6t+62

    models the percentage of jobs in the U.S. labor force, p, held by men t years after 1970.

    1. Use the appropriate graph to estimate the percentage of jobs in the U.S. labor force held by men in 2010. Give your estimation to the nearest percent.

    2. Use the mathematical model to determine the percentage of jobs in the U.S. labor force held by men in 2010. Round to the nearest tenth of a percent.

    3. According to the formula, when will 49% of jobs in the U.S. labor force be held by men? Round to the nearest year.

Explaining the Concepts

  1. 145. What is a linear equation in one variable? Give an example of this type of equation.

  2. 146. Explain how to determine the restrictions on the variable for the equation

    3x+5+4x2=7x2+3x6.
  3. 147. What does it mean to solve a formula for a variable?

  4. 148. Explain how to solve an equation involving absolute value.

  5. 149. Why does the procedure that you explained in Exercise 148 not apply to the equation |x2|=3? What is the solution set for this equation?

  6. 150. What is a quadratic equation?

  7. 151. Explain how to solve x2+6x+8=0 using factoring and the zero-product principle.

  8. 152. Explain how to solve x2+6x+8=0 by completing the square.

  9. 153. Explain how to solve x2+6x+8=0 using the quadratic formula.

  10. 154. How is the quadratic formula derived?

  11. 155. What is the discriminant and what information does it provide about a quadratic equation?

  12. 156. If you are given a quadratic equation, how do you determine which method to use to solve it?

  13. 157. In solving 2x1+2=x, why is it a good idea to isolate the radical term? What if we don’t do this and simply square each side? Describe what happens.

  14. 158. What is an extraneous solution to a radical equation?

Critical Thinking Exercises

Make Sense? In Exercises 159162, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 159. The model p=0.8x+25 describes the percentage of college freshmen with an A average in high school, p, x years after 1980, so I have to solve a linear equation to determine the percentage of college freshmen with an A average in high school in 2020.

  2. 160. Although I can solve 3x+15=14 by first subtracting 15 from both sides, I find it easier to begin by multiplying both sides by 20, the least common denominator.

  3. 161. Because I want to solve 25x2169=0 fairly quickly, I’ll use the quadratic formula.

  4. 162. When checking a radical equation’s proposed solution, I can substitute into the original equation or any equation that is part of the solution process.

In Exercises 163166, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 163. The equation (2x3)2=25 is equivalent to 2x3=5.

  2. 164. Every quadratic equation has two distinct numbers in its solution set.

  3. 165. The equations 3y1=11 and 3y7=5 are equivalent.

  4. 166. The equation ax2+c=0, a0, cannot be solved by the quadratic formula.

  5. 167. Find b such that 7x+4b+13=x will have a solution set given by {6}.

  6. 168. Write a quadratic equation in standard form whose solution set is {3, 5}.

  7. 169. Solve for C:V=CCSL N.

  8. 170. Solve for t:s=16t2+v0t.

Preview Exercises

Exercises 171173 will help you prepare for the material covered in the next section.

  1. 171. Jane’s salary exceeds Jim’s by $150 per week. If x represents Jim’s weekly salary, write an algebraic expression that models Jane’s weekly salary.

  2. 172. A convenience store sells a refillable stainless-steel travel mug for $20. Customers who buy the mug can refill it with coffee for just $0.99 on each visit. Write an algebraic expression that models the total cost of the mug and x refills.

  3. 173. If the width of a rectangle is represented by x and the length is represented by x+200, write a simplified algebraic expression that models the rectangle’s perimeter.

P.7: Exercise Set

P.7 Exercise Set

Practice Exercises

In Exercises 116, solve and check each linear equation.

  1. 1. 7x5=72

  2. 2. 6x3=63

  3. 3. 11x(6x5)=40

  4. 4. 5x(2x10)=35

  5. 5. 2x7=6+x

  6. 6. 3x+5=2x+13

  7. 7. 7x+4=x+16

  8. 8. 13x+14=12x5

  9. 9. 3(x2)+7=2(x+5)

  10. 10. 2(x1)+3=x3(x+1)

  11. 11. x+36=38+x54

  12. 12. x+14=16+2x3

  13. 13. x4=2+x33

  14. 14. 5+x23=x+38

  15. 15. x+13=5x+27

  16. 16. 3x5x32=x+23

Exercises 1726 contain rational equations with variables in denominators. For each equation, a. Write the value or values of the variable that make a denominator zero. These are the restrictions on the variable. b. Keeping the restrictions in mind, solve the equation.

  1. 17. 1x1+5=11x1

  2. 18. 3x+47=4x+4

  3. 19. 8xx+1=48x+1

  4. 20. 2x2=xx22

  5. 21. 32x2+12=2x1

  6. 22. 3x+3=52x+6+1x2

  7. 23. 2x+11x1=2xx21

  8. 24. 4x+5+2x5=32x225

  9. 25. 1x45x+2=6x22x8

  10. 26. 1x32x+1=8x22x3

In Exercises 2742, solve each formula for the specified variable. Do you recognize the formula? If so, what does it describe?

  1. 27. I=Prt for P

  2. 28. C=2πr for r

  3. 29. T=D+pm for p

  4. 30. P=C+MC for M

  5. 31. A=12h(a+b) for a

  6. 32. A=12h(a+b) for b

  7. 33. S=P+Prt for r

  8. 34. S=P+Prt for t

  9. 35. B=FSV for S

  10. 36. S=C1r for r

  11. 37. IR+Ir=E for I

  12. 38. A=2lw+2lh+2wh for h

  13. 39. 1p+1q=1f for f

  14. 40. 1R=1R1+1R2 for R1

  15. 41. f=f1f2f1+f2 for f1

  16. 42. f=f1f2f1+f2 for f2

In Exercises 4354, solve each absolute value equation or indicate the equation has no solution.

  1. 43. |x2|=7

  2. 44. |x+1|=5

  3. 45. |2x1|=5

  4. 46. |2x3|=11

  5. 47. 2|3x2|=14

  6. 48. 3|2x1|=21

  7. 49. 2|452 x|+6=18

  8. 50. 4|134 x|+7=10

  9. 51. |x+1|+5=3

  10. 52. |x+1|+6=2

  11. 53. |2x1|+3=3

  12. 54. |3x2|+4=4

In Exercises 5560, solve each quadratic equation by factoring.

  1. 55. x23x10=0

  2. 56. x213x+36=0

  3. 57. x2=8x15

  4. 58. x2=11x10

  5. 59. 5x2=20x

  6. 60. 3x2=12x

In Exercises 6166, solve each quadratic equation by the square root property.

  1. 61. 3x2=27

  2. 62. 5x2=45

  3. 63. 5x2+1=51

  4. 64. 3x21=47

  5. 65. 3(x4)2=15

  6. 66. 3(x+4)2=21

In Exercises 6774, solve each quadratic equation by completing the square.

  1. 67. x2+6x=7

  2. 68. x2+6x=8

  3. 69. x22x=2

  4. 70. x2+4x=12

  5. 71. x26x11=0

  6. 72. x22x5=0

  7. 73. x2+4x+1=0

  8. 74. x2+6x5=0

In Exercises 7582, solve each quadratic equation using the quadratic formula.

  1. 75. x2+8x+15=0

  2. 76. x2+8x+12=0

  3. 77. x2+5x+3=0

  4. 78. x2+5x+2=0

  5. 79. 3x23x4=0

  6. 80. 5x2+x2=0

  7. 81. 4x2=2x+7

  8. 82. 3x2=6x1

Compute the discriminant of each equation in Exercises 8390. What does the discriminant indicate about the number and type of solutions?

  1. 83. x24x5=0

  2. 84. 4x22x+3=0

  3. 85. 2x211x+3=0

  4. 86. 2x2+11x6=0

  5. 87. x2=2x1

  6. 88. 3x2=2x1

  7. 89. x23x7=0

  8. 90. 3x2+4x2=0

In Exercises 91114, solve each quadratic equation by the method of your choice.

  1. 91. 2x2x=1

  2. 92. 3x24x=4

  3. 93. 5x2+2=11x

  4. 94. 5x2=613x

  5. 95. 3x2=60

  6. 96. 2x2=250

  7. 97. x22x=1

  8. 98. 2x2+3x=1

  9. 99. (2x+3)(x+4)=1

  10. 100. (2x5)(x+1)=2

  11. 101. (3x4)2=16

  12. 102. (2x+7)2=25

  13. 103. 3x212x+12=0

  14. 104. 96x+x2=0

  15. 105. 4x216=0

  16. 106. 3x227=0

  17. 107. x2=4x2

  18. 108. x2=6x7

  19. 109. 2x27x=0

  20. 110. 2x2+5x=3

  21. 111. 1x+1x+2=13

  22. 112. 1x+1x+3=14

  23. 113. 2xx3+6x+3=28x29

  24. 114. 3x3+5x4=x220x27x+12

In Exercises 115124, solve each radical equation. Check all proposed solutions.

  1. 115. 3x+18=x

  2. 116. 208x=x

  3. 117. x+3=x3

  4. 118. x+10=x2

  5. 119. 2x+13=x+7

  6. 120. 6x+1=x1

  7. 121. x2x+5=5

  8. 122. xx+11=1

  9. 123. 2x+198=x

  10. 124. 2x+156=x

Practice PLUS

In Exercises 125134, solve each equation.

  1. 125. 25[2+5x3(x+2)]=3(2x5)[5(x1)3x+3]

  2. 126. 45[42x4(x+7)]=4(1+3x)[43(x+2)2(2x5)]

  3. 127. 77x=(3x+2)(x1)

  4. 128. 10x1=(2x+1)2

  5. 129. |x2+2x36|=12

  6. 130. |x2+6x+1|=8

  7. 131. 1x23x+2=1x+2+5x24

  8. 132. x1x2+xx3=1x25x+6

  9. 133. x+8x4=2

  10. 134. x+5x3=2

In Exercises 135136, list all numbers that must be excluded from the domain of each rational expression.

  1. 135. 32x2+4x9

  2. 136. 72x28x+5

Application Exercises

Grade Inflation. The bar graph shows the percentage of U.S. college freshmen with an average grade of A in high school.

A bar graph depicts the percentage of U.S. college freshmen with an average grade of A left parenthesis A negative to A positive right parenthesis in high school.

Source: Higher Education Research Institute

The data displayed by the bar graph can be described by the mathematical model

p=4x5+25,

where x is the number of years after 1980 and p is the percentage of U.S. college freshmen who had an average grade of A in high school. Use this information to solve Exercises 137138.

  1. 137.

    1. According to the formula, in 2010, what percentage of U.S. college freshmen had an average grade of A in high school? Does this underestimate or overestimate the percent displayed by the bar graph? By how much?

    2. If trends shown by the formula continue, project when 57% of U.S. college freshmen will have had an average grade of A in high school.

  2. 138.

    1. According to the formula, in 2000, what percentage of U.S. college freshmen had an average grade of A in high school? Does this underestimate or overestimate the percent displayed by the bar graph? By how much?

    2. If trends shown by the formula continue, project when 65% of U.S. college freshmen will have had an average grade of A in high school.

  3. 139. A company wants to increase the 10% peroxide content of its product by adding pure peroxide (100% peroxide). If x liters of pure peroxide are added to 500 liters of its 10% solution, the concentration, C, of the new mixture is given by

    C=x+0.1(500)x+500.

    How many liters of pure peroxide should be added to produce a new product that is 28% peroxide?

  4. 140. Suppose that x liters of pure acid are added to 200 liters of a 35% acid solution.

    1. Write a formula that gives the concentration, C, of the new mixture. (Hint: See Exercise 139.)

    2. How many liters of pure acid should be added to produce a new mixture that is 74% acid?

A driver’s age has something to do with his or her chance of getting into a fatal car crash. The bar graph shows the number of fatal vehicle crashes per 100 million miles driven for drivers of various age groups. For example, 25-year-old drivers are involved in 4.1 fatal crashes per 100 million miles driven. Thus, when a group of 25-year-old Americans have driven a total of 100 million miles, approximately 4 have been in accidents in which someone died.

A vertical bar graph titled, Age of United States drivers and fatal crashes.

Source: Insurance Institute for Highway Safety

The number of fatal vehicle crashes per 100 million miles, N, for drivers of age x can be modeled by the formula

N=0.013x21.19x+28.24.

Use the formula to solve Exercises 141142.

  1. 141. What age groups are expected to be involved in 3 fatal crashes per 100 million miles driven? How well does the formula model the trend in the actual data shown in the bar graph?

  2. 142. What age groups are expected to be involved in 10 fatal crashes per 100 million miles driven? How well does the formula model the trend in the actual data shown in the bar graph?

The graphs show the percentage of jobs in the U.S. labor force held by men and by women from 1970 through 2015. Exercises 143144 are based on the data displayed by the graphs.

The line graph titled, Percentage of U.S Jobs held by Men and Women.

Source: Bureau of Labor Statistics

  1. 143. The formula

    p=1.6t+38

    models the percentage of jobs in the U.S. labor force, p, held by women t years after 1970.

    1. Use the appropriate graph to estimate the percentage of jobs in the U.S. labor force held by women in 2010. Give your estimation to the nearest percent.

    2. Use the mathematical model to determine the percentage of jobs in the U.S. labor force held by women in 2010. Round to the nearest tenth of a percent.

    3. According to the formula, when will 51% of jobs in the U.S. labor force be held by women? Round to the nearest year.

  2. 144. The formula

    p=1.6t+62

    models the percentage of jobs in the U.S. labor force, p, held by men t years after 1970.

    1. Use the appropriate graph to estimate the percentage of jobs in the U.S. labor force held by men in 2010. Give your estimation to the nearest percent.

    2. Use the mathematical model to determine the percentage of jobs in the U.S. labor force held by men in 2010. Round to the nearest tenth of a percent.

    3. According to the formula, when will 49% of jobs in the U.S. labor force be held by men? Round to the nearest year.

Explaining the Concepts

  1. 145. What is a linear equation in one variable? Give an example of this type of equation.

  2. 146. Explain how to determine the restrictions on the variable for the equation

    3x+5+4x2=7x2+3x6.
  3. 147. What does it mean to solve a formula for a variable?

  4. 148. Explain how to solve an equation involving absolute value.

  5. 149. Why does the procedure that you explained in Exercise 148 not apply to the equation |x2|=3? What is the solution set for this equation?

  6. 150. What is a quadratic equation?

  7. 151. Explain how to solve x2+6x+8=0 using factoring and the zero-product principle.

  8. 152. Explain how to solve x2+6x+8=0 by completing the square.

  9. 153. Explain how to solve x2+6x+8=0 using the quadratic formula.

  10. 154. How is the quadratic formula derived?

  11. 155. What is the discriminant and what information does it provide about a quadratic equation?

  12. 156. If you are given a quadratic equation, how do you determine which method to use to solve it?

  13. 157. In solving 2x1+2=x, why is it a good idea to isolate the radical term? What if we don’t do this and simply square each side? Describe what happens.

  14. 158. What is an extraneous solution to a radical equation?

Critical Thinking Exercises

Make Sense? In Exercises 159162, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 159. The model p=0.8x+25 describes the percentage of college freshmen with an A average in high school, p, x years after 1980, so I have to solve a linear equation to determine the percentage of college freshmen with an A average in high school in 2020.

  2. 160. Although I can solve 3x+15=14 by first subtracting 15 from both sides, I find it easier to begin by multiplying both sides by 20, the least common denominator.

  3. 161. Because I want to solve 25x2169=0 fairly quickly, I’ll use the quadratic formula.

  4. 162. When checking a radical equation’s proposed solution, I can substitute into the original equation or any equation that is part of the solution process.

In Exercises 163166, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 163. The equation (2x3)2=25 is equivalent to 2x3=5.

  2. 164. Every quadratic equation has two distinct numbers in its solution set.

  3. 165. The equations 3y1=11 and 3y7=5 are equivalent.

  4. 166. The equation ax2+c=0, a0, cannot be solved by the quadratic formula.

  5. 167. Find b such that 7x+4b+13=x will have a solution set given by {6}.

  6. 168. Write a quadratic equation in standard form whose solution set is {3, 5}.

  7. 169. Solve for C:V=CCSL N.

  8. 170. Solve for t:s=16t2+v0t.

Preview Exercises

Exercises 171173 will help you prepare for the material covered in the next section.

  1. 171. Jane’s salary exceeds Jim’s by $150 per week. If x represents Jim’s weekly salary, write an algebraic expression that models Jane’s weekly salary.

  2. 172. A convenience store sells a refillable stainless-steel travel mug for $20. Customers who buy the mug can refill it with coffee for just $0.99 on each visit. Write an algebraic expression that models the total cost of the mug and x refills.

  3. 173. If the width of a rectangle is represented by x and the length is represented by x+200, write a simplified algebraic expression that models the rectangle’s perimeter.

Objective 1: Use equations to solve problems

Problem Solving with Equations

  1. Objective 1Use equations to solve problems.

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We have seen that a model is a mathematical representation of a real-world situation. In this section, we will be solving problems that are presented in English. This means that we must obtain models by translating from the ordinary language of English into the language of algebraic equations. To translate, however, we must understand the English prose and be familiar with the forms of algebraic language. Following are some general steps we will use in solving word problems:

Strategy for Solving Word Problems

  1. Step 1 Read the problem carefully several times until you can state in your own words what is given and what the problem is looking for. Let x (or any variable) represent one of the unknown quantities in the problem.

  2. Step 2 If necessary, write expressions for any other unknown quantities in the problem in terms of x.

  3. Step 3 Write an equation in x that models the verbal conditions of the problem.

  4. Step 4 Solve the equation and answer the problem’s question.

  5. Step 5 Check the solution in the original wording of the problem, not in the equation obtained from the words.

Example 1 Education Pays Off

The graph in Figure P.14 shows median yearly earnings in the United States by highest educational attainment. It certainly looks like you can make more money with more education. However, workers who make more money pay a higher percentage of income in taxes, so it is wise to consider after-tax earnings when deciding if that higher degree will pay off.

Figure P.14

A bar chart shows median earnings of full-time workers in the United States, by highest educational attainment.

Source: Education Pays 2019

The median yearly after-tax income of a full-time worker with a bachelor’s degree exceeds that of a full-time worker with an associate’s degree by $11 thousand. The median yearly after-tax income of a full-time worker with a master’s degree exceeds that of a full-time worker with an associate’s degree by $21 thousand. Combined, three full-time workers with each of these degrees earn $149 thousand after taxes. Find the median yearly after-tax income of full-time workers with each of these levels of education.

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We know something about after-tax incomes of full-time workers with bachelor’s degrees and master’s degrees: They exceed the after-tax income of a full-time worker with an associate’s degree by $11 thousand and $21 thousand, respectively. We will let

    x=the median yearly after-tax income of a full-time worker with an associates degree.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. Because a full-time worker with a bachelor’s degree earns $11 thousand more after taxes than a full-time worker with an associate’s degree, let

    x+11=the median yearly after-tax income of a full-time worker with a bachelors degree.

    Because a full-time worker with a master’s degree earns $21 thousand more after taxes than a full-time worker with an associate’s degree, let

    x+21=the median yearly after-tax income of a full-time worker with a master's degree.
  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. Combined, three full-time workers with each of these degrees earn $149 thousand.

    An equation with variable x to model the conditions.

  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    x+(x+11)+(x+21)=149This is the equation that models the problems conditions.3x+32=149Remove parentheses, regroup, and combine like terms.3x=117Subtract 32 from both sides.x=39Divide both sides by 3.

    Because we isolated the variable in the model and obtained x=39,

    median after-tax income with an associates degree=x=39median after-tax income with a bachelors degree=x+11=39+11=50median after-tax income with a masters degree=x+21=39+21=60.

    Full-time workers with associate’s degrees earn $39 thousand per year after taxes, full-time workers with bachelor’s degrees earn $50 thousand per year after taxes, and full-time workers with master’s degrees earn $60 thousand per year after taxes.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that combined, three full-time workers with each of these educational attainments earn $149 thousand after taxes. Using the incomes we determined in step 4, the sum is

    $39 thousand+$50 thousand+$60 thousand,or $149 thousand,

    which satisfies the problem’s conditions.

Check Point 1

  • The median yearly before-tax income of a full-time worker with a bachelor’s degree exceeds that of a full-time worker with an associate’s degree by $15 thousand. The median yearly before-tax income of a full-time worker with a master’s degree exceeds that of a full-time worker with an associate’s degree by $30 thousand. Combined, three full-time workers with each of these educational attainments earn $195 thousand before taxes. Find the median yearly before-tax income of full-time workers with each of these levels of education. (These incomes are illustrated by the bar graph on the previous page.)

Example 2 Modeling Attitudes of College Freshmen

Your author teaching math in 1969

Researchers have surveyed college freshmen every year since 1969. Figure P.15 shows that attitudes about some life goals have changed dramatically over the years. In particular, the freshman class of 2018 was more interested in making money than the freshmen of 1969 had been. In 1969, 42% of first-year college students considered “being well-off financially” essential or very important. For the period from 1969 through 2018, this percentage increased by approximately 0.8 each year. If this trend continues, by which year will all college freshmen consider “being well-off financially” essential or very important?

Figure P.15

A bar chart shows the life objectives of college freshmen, for the period from 19 69 to 20 18.

Source: Higher Education Research Institute

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We are interested in the year when all college freshmen, or 100% of the freshmen, will consider this life objective essential or very important. Let

    x = the number of years after 1969 when all freshmen will considerbeing    well-off financially essential or very important.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. There are no other unknown quantities to find, so we can skip this step.

  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS.

    The image shows a mathematical expression 42 + 0.8 x = 100
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    42+0.8x=100This is the equation that models the problems conditions.4242+0.8x=10042Subtract 42 from both sides.0.8x=58Simplify.0.8x0.8=580.8Divide both sides by 0.8.x=72.573Simplify and round to the nearestwhole number.

    Using current trends, by approximately 73 years after 1969, or in 2042, all freshmen will consider “being well-off financially” essential or very important.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that all freshmen (100%, represented by 100 using the model) will consider the objective essential or very important. Does this approximately occur if we increase the 1969 percentage, 42, by 0.8 each year for 73 years, our proposed solution?

    42+0.8(73)=42+58.4=100.4100

    This verifies that using trends shown in Figure P.15, all first-year college students will consider the objective of being well-off financially essential or very important approximately 73 years after 1969, or in 2042.

Check Point 2

  • Figure P.15 on the previous page shows that the freshman class of 2018 was less interested in developing a philosophy of life than the freshmen of 1969 had been. In 1969, 85% of the freshmen considered this objective essential or very important. Since then, this percentage has decreased by approximately 0.8 each year. If this trend continues, by which year will only 25% of college freshmen consider “developing a meaningful philosophy of life” essential or very important?

Example 3 Modeling Options for a Toll

It’s time to decide how you’re going to pay the bridge toll so that you can get to the beach. The first option requires purchasing a transponder for $20; with the transponder, you pay a reduced toll of $3.25 each time you cross the bridge. Your second option is toll-by-plate; with this option, you pay the full toll of $4.25 each time you cross the bridge plus a $3 administrative fee for each crossing. Find the number of times you would need to cross the bridge for the costs of the two options to be the same.

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. Let

    x=the number of times you cross the bridge.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. There are no other unknown quantities, so we can skip this step.

  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. The cost with the transponder is the cost of the transponder, $20, plus the toll, $3.25, times the number of crossings, x. For toll-by-plate, you must pay both the toll, $4.25, and the administrative fee, $3, each time you cross the bridge. The cost for toll-by-plate is the sum of the toll and the administrative fee, $4.25+$3 or $7.25, times the number of crossings, x. We want to know when these two costs are equal.

    The image shows a mathematical expression 7.25x = 20 + 3.25 x
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    7.25x=20+3.25xThis is the equation that models theproblems conditions.4x=20Subtract 3.25x from both sides.4x=5Divide both sides by 4.

    Because x represents the number of times you cross the bridge, the total cost of toll-by-plate is the same as the total cost with the transponder for 5 crossings.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that the cost of toll-by-plate should be the same as the cost with the transponder. Let’s see if they are the same with 5 bridge crossings per month.

    Expression to check the proposed solution.

    If you cross the bridge five times, both options cost $36.25. Thus the proposed solution, 5 crossings, satisfies the problem’s conditions.

Check Point 3

  • You drive up to a toll plaza and find booths with attendants where you can pay the toll by cash or credit card. With this option, the toll is $5 each time you cross the bridge. The attendant gives you the option of buying a bar-coded decal for $25; with the decal, you get 25% off the normal toll of $5 for each crossing. Find the number of times you would need to cross the bridge for the costs of the two options to be the same.

Example 4 A Price Reduction on Wireless Headphones

Your favorite electronics retailer is having a terrific sale on noise cancelling wireless headphones. (You’ll finally be able to listen to music in peace!) After a 40% price reduction, you purchase headphones for $192. What was the price of the headphones before the reduction?

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the price of the headphones prior to the reduction.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. There are no other unknown quantities to find, so we can skip this step.

  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. The headphones’ original price minus the 40% reduction is the reduced price, $192.

    The image shows a mathematical expression x minus 0.4 x = 192.
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    The solution of an equation.

    The headphones’ price before the reduction was $320.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The price before the reduction, $320, minus the 40% reduction should equal the reduced price given in the original wording, $192:

    32040% of 320=3200.4(320)=320128=192.

    This verifies that the headphones’ price before the reduction was $320.

Check Point 4

  • After a 30% price reduction, you purchase a new laptop for $840. What was the laptop’s price before the reduction?

Our next example is about simple interest. Simple interest involves interest calculated only on the amount of money that we invest, called the principal. The formula I=Pr is used to find the simple interest, I, earned for one year when the principal, P, is invested at an annual interest rate, r. Dual investment problems involve different amounts of money in two or more investments, each paying a different rate.

Example 5 Solving a Dual Investment Problem

Your grandmother needs your help. She has $150,000 to invest. Part of this money is to be invested in a money market account paying 1.5% annual interest. The rest of this money is to be invested in a certificate of deposit paying 1.3% annual interest. She told you that she requires $2000 per year in extra income from the combination of these investments. How much money should be placed in each investment?

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the amount invested in the money market account at 1.5%.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. The other quantity that we seek is the amount invested at 1.3% in the certificate of deposit. Because the total amount Grandma has to invest is $150,000 and we already used up x,

    150,000x=the amount invested in the certificate of deposit at 1.3%.
  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. Because Grandma requires $2000 in total interest, the interest for the two investments combined must be $2000. Interest is Pr or rP for each investment.

    The image shows a mathematical expression 0.015 x + 0.013 left parenthesis 150,000 minus x right parenthesis = 20 00
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    A mathematical expression steps to solve an equation to find the value of x.

    Thus,

    the amount invested at1.5%=x=25,000the amount invested at 1.3%=150,00025,000=125,000

    Grandma should invest $25,000 at 1.5% and $125,000 at 1.3%.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that the total interest from the dual investments should be $2000. Can Grandma count on $2000 interest? The interest earned on $25,000 at 1.5% is ($25,000) (0.015), or $375. The interest earned on $125,000 at 1.3% is ($125,000)(0.013), or $1625. The total interest is $375+$1625, or $2000, exactly as it should be. You’ve made your grandmother happy. (Now if you would just visit her more often …)

Check Point 5

  • You inherited $50,000 with the stipulation that for the first year the money had to be invested in two accounts paying 0.9% and 1.1% annual interest. How much did you invest at each rate if the total interest earned for the year was $515?

Solving geometry problems usually requires a knowledge of basic geometric ideas and formulas. Formulas for area, perimeter, and volume are given in Table P.6.

Table P.6 Common Formulas for Area, Perimeter, and Volume

An illustration shows common formulas for area, perimeter, and volume of different shapes.
Table P.6 Full Alternative Text

We will be using the formula for the perimeter of a rectangle, P=2l+2w, in our next example. The formula states that a rectangle’s perimeter is the sum of twice its length and twice its width.

Example 6 Finding the Dimensions of an American Football Field

The length of an American football field is 200 feet more than the width. If the perimeter of the field is 1040 feet, what are its dimensions?

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We know something about the length; the length is 200 feet more than the width. We will let

    x=the width.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. Because the length is 200 feet more than the width, we add 200 to the width to represent the length. Thus,

    x+200=the length.

    Figure P.16 illustrates an American football field and its dimensions.

    Figure P.16 An American football field

    An illustration shows an American football field with its width as x and length as x + 20 0.
  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. Because the perimeter of the field is 1040 feet,

    The image shows the mathematical expression 2 left parenthesis x + 20 0 right parenthesis + 2 x = 1040.

  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    Steps to solve an equation.

    Thus,

    width=x=160.length=x+200=160+200=360.

    The dimensions of an American football field are 160 feet by 360 feet. (The 360-foot length is usually described as 120 yards.)

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The perimeter of the football field using the dimensions that we found is

    2(160 feet)+2(360 feet)=320 feet+720 feet=1040 feet.

    Because the problem’s wording tells us that the perimeter is 1040 feet, our dimensions are correct.

Check Point 6

  • The length of a rectangular basketball court is 44 feet more than the width. If the perimeter of the basketball court is 288 feet, what are its dimensions?

We will use the formula for the area of a rectangle, A=lw, in our next example. The formula states that a rectangle’s area is the product of its length and its width.

Example 7 Solving a Problem Involving Landscape Design

A rectangular garden measures 80 feet by 60 feet. A large path of uniform width is to be added along both shorter sides and one longer side of the garden. The landscape designer doing the work wants to double the garden’s area with the addition of this path. How wide should the path be?

Solution

  1. Step 1 LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the width of the path.

    The situation is illustrated in Figure P.17. The figure shows the original 80-by-60 foot rectangular garden and the path of width x added along both shorter sides and one longer side.

    Figure P.17 The garden’s area is to be doubled by adding the path.

    An image shows a rectangular garden is surrounded by a path that has a length of 80 + 2 x and a width of 60 + x.
  2. Step 2 REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. Because the path is added along both shorter sides and one longer side, Figure P.17 shows that

     80+2x=the length of the new, expanded rectangle 60+x=the width of the new, expanded rectangle.

  3. Step 3 WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. The area of the rectangle must be doubled by the addition of the path.

    Left parenthesis 80 + 2 x right parenthesis, left parenthesis 60 + x right parenthesis = 2 times 80 times 60.
  4. Step 4 SOLVE THE EQUATION AND ANSWER THE QUESTION.

    (80+2x)(60+x)=28060This is the equation that modelsthe problems conditions.4800+200x+2x2=9600Multiply. Use FOIL on the left side.2x2+200x4800=0Subtract 9600 from both sidesand write the quadratic equationin standard form.2(x2+100x2400)=0Factor out 2, the GCF.2(x20)(x+120)=0Factor the trinomial.x20=0or x+120=0Set each variable factor equal to 0.x=20x=120Solve for x.

    The path cannot have a negative width. Because 120 is geometrically impossible, we use x=20. The width of the path should be 20 feet.

  5. Step 5 CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. Has the landscape architect doubled the garden’s area with the 20-foot-wide path? The area of the garden is 80 feet times 60 feet, or 4800 square feet. Because 80+2x and 60+x represent the length and width of the expanded rectangle,

    80+2x=80+220 =120 feet is the expanded rectangle's length.60+2x=60+20=80 feet is the expanded rectangle's width.

    The area of the expanded rectangle is 120 feet times 80 feet, or 9600 square feet. This is double the area of the garden, 4800 square feet, as specified by the problem’s conditions.

Check Point 7

  • A rectangular garden measures 16 feet by 12 feet. A path of uniform width is to be added so as to surround the entire garden. The landscape artist doing the work wants the garden and path to cover an area of 320 square feet. How wide should the path be?

In our next example, we will be using the Pythagorean Theorem to obtain a mathematical model. The ancient Greek philosopher and mathematician Pythagoras (approximately 582–500 b.c.) founded a school whose motto was “All is number.” Pythagoras is best remembered for his work with the right triangle, a triangle with one angle measuring 90°. The side opposite the 90° angle is called the hypotenuse. The other sides are called legs. Pythagoras found that if he constructed squares on each of the legs, as well as a larger square on the hypotenuse, the sum of the areas of the smaller squares is equal to the area of the larger square. This is illustrated in Figure P.18.

Figure P.18The area of the large square equals the sum of the areas of the smaller squares.

An illustration shows a right triangle of base 3, height 4, and hypotenuse 5.
Figure P. 18 Full Alternative Text

This relationship is usually stated in terms of the lengths of the three sides of a right triangle and is called the Pythagorean Theorem.

The Pythagorean Theorem

The sum of the squares of the lengths of the legs of a right triangle equals the square of the length of the hypotenuse.

If the legs have lengths a and b, and the hypotenuse has length c, then

a2+b2=c2.
An image shows a right angle triangle ABC, which represents angle C is a 90 degree angle. AC represents as a base labeled, b, Leg, BC denotes as a height labeled, a Leg, and AB is labeled, c, hypotenuse.

Example 8 Using the Pythagorean Theorem: Screen Math

A bar graph shows the rising percentage of large screen TVs on store shelves.

Source: GAP Intelligence

Did you know that the size of a television screen refers to the length of its diagonal? If the length of the HDTV screen in Figure P.19 is 28 inches and its width is 15.7 inches, what is the size of the screen to the nearest inch?

Figure P.19

A photo shows a television with a screen size of length 28 inches and a width of 15.7 inches.

Solution

Figure P.19 shows that the length, width, and diagonal of the screen form a right triangle. The diagonal is the hypotenuse of the triangle. We use the Pythagorean Theorem with a=28, b=15.7, and solve for the screen size, c.

a2+b2=c2This is the Pythagorean Theorem.282+15.72=c2Let a=28 and b=15.7.784+246.49=c2282=2828=784 and15.72=15.715.7=246.49.c2=1030.49Add: 784+246.49=1030.49. We also reversedthe two sides.
c=1030.49 or  c=1030.49Apply the square root property.c32 or  c32Use a calculator and round to thenearest inch. 1030.49    = or   1030.49 ENTER

Because c represents the size of the screen, this dimension must be positive. We reject 32. Thus, the screen size of the HDTV is 32 inches.

Check Point 8

  • Figure P.20 shows the dimensions of an old TV screen. What is the size of the screen?

    Figure P.20

    A photo shows a television with a screen size of length 25.6 inches and a width of 19.2 inches.

Example 9 Dividing the Cost of a Yacht

A group of friends agrees to share the cost of a $50,000 yacht equally. Before the purchase is made, one more person joins the group. As a result, each person’s share is reduced by $2500. How many people were in the original group?

Solution

  1. Step 1 LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the number of people in the original group.
  2. Step 2 REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x.  Because one more person joined the original group, let

    x+1=the number of people in the final group.
  3. Step 3 WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. As a result of one more person joining the original group, each person’s share is reduced by $2500.

    The image shows a mathematical expression, start fraction 50,000 over x end fraction minus 2500 = start fraction 50,000 over x + 1 end fraction.
  4. Step 4 SOLVE THE EQUATION AND ANSWER THE QUESTION.

    50,000x2500=50,000x+1This is the equation that models theproblems conditions.x(x+1)(50,000x2500)=x(x+1)50,000x+1Multiply both sides by x(x+1),the LCD. x (x+1)50,000 x x(x+1)2500=x (x+1) 50,000 (x+1) Use the distributive property anddivide out common factors.50,000(x+1)2500x(x+1)=50,000xSimplify.50,000x+50,0002500x22500x=50,000xUse the distributive property.2500x2+47,500x+50,000=50,000xCombine like terms.2500x22500x+50,000=0Write the quadratic equation instandard form.2500(x2+x20)=0Factor out 2500.2500(x+5)(x4)=0Factor completely.x+5=0or x4=0Set each variable factor equal to zero.x=5x=4Solve the resulting equations.

    Because x represents the number of people in the original group, x cannot be negative. Thus, there were four people in the original group.

  5. Step 5 CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM.

     original cost per person=$50,0004=$12,500 final cost per person=$50,0005=$10,000

    We see that each person’s share is reduced by $12,500$10,000, or $2500, as specified by the problem’s conditions.

Check Point 9

  • A group of people share equally in a $5,000,000 lottery. Before the money is divided, three more winning ticket holders are declared. As a result, each person’s share is reduced by $375,000. How many people were in the original group of winners?

Objective 1: Use equations to solve problems

Problem Solving with Equations

  1. Objective 1Use equations to solve problems.

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We have seen that a model is a mathematical representation of a real-world situation. In this section, we will be solving problems that are presented in English. This means that we must obtain models by translating from the ordinary language of English into the language of algebraic equations. To translate, however, we must understand the English prose and be familiar with the forms of algebraic language. Following are some general steps we will use in solving word problems:

Strategy for Solving Word Problems

  1. Step 1 Read the problem carefully several times until you can state in your own words what is given and what the problem is looking for. Let x (or any variable) represent one of the unknown quantities in the problem.

  2. Step 2 If necessary, write expressions for any other unknown quantities in the problem in terms of x.

  3. Step 3 Write an equation in x that models the verbal conditions of the problem.

  4. Step 4 Solve the equation and answer the problem’s question.

  5. Step 5 Check the solution in the original wording of the problem, not in the equation obtained from the words.

Example 1 Education Pays Off

The graph in Figure P.14 shows median yearly earnings in the United States by highest educational attainment. It certainly looks like you can make more money with more education. However, workers who make more money pay a higher percentage of income in taxes, so it is wise to consider after-tax earnings when deciding if that higher degree will pay off.

Figure P.14

A bar chart shows median earnings of full-time workers in the United States, by highest educational attainment.

Source: Education Pays 2019

The median yearly after-tax income of a full-time worker with a bachelor’s degree exceeds that of a full-time worker with an associate’s degree by $11 thousand. The median yearly after-tax income of a full-time worker with a master’s degree exceeds that of a full-time worker with an associate’s degree by $21 thousand. Combined, three full-time workers with each of these degrees earn $149 thousand after taxes. Find the median yearly after-tax income of full-time workers with each of these levels of education.

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We know something about after-tax incomes of full-time workers with bachelor’s degrees and master’s degrees: They exceed the after-tax income of a full-time worker with an associate’s degree by $11 thousand and $21 thousand, respectively. We will let

    x=the median yearly after-tax income of a full-time worker with an associates degree.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. Because a full-time worker with a bachelor’s degree earns $11 thousand more after taxes than a full-time worker with an associate’s degree, let

    x+11=the median yearly after-tax income of a full-time worker with a bachelors degree.

    Because a full-time worker with a master’s degree earns $21 thousand more after taxes than a full-time worker with an associate’s degree, let

    x+21=the median yearly after-tax income of a full-time worker with a master's degree.
  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. Combined, three full-time workers with each of these degrees earn $149 thousand.

    An equation with variable x to model the conditions.

  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    x+(x+11)+(x+21)=149This is the equation that models the problems conditions.3x+32=149Remove parentheses, regroup, and combine like terms.3x=117Subtract 32 from both sides.x=39Divide both sides by 3.

    Because we isolated the variable in the model and obtained x=39,

    median after-tax income with an associates degree=x=39median after-tax income with a bachelors degree=x+11=39+11=50median after-tax income with a masters degree=x+21=39+21=60.

    Full-time workers with associate’s degrees earn $39 thousand per year after taxes, full-time workers with bachelor’s degrees earn $50 thousand per year after taxes, and full-time workers with master’s degrees earn $60 thousand per year after taxes.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that combined, three full-time workers with each of these educational attainments earn $149 thousand after taxes. Using the incomes we determined in step 4, the sum is

    $39 thousand+$50 thousand+$60 thousand,or $149 thousand,

    which satisfies the problem’s conditions.

Check Point 1

  • The median yearly before-tax income of a full-time worker with a bachelor’s degree exceeds that of a full-time worker with an associate’s degree by $15 thousand. The median yearly before-tax income of a full-time worker with a master’s degree exceeds that of a full-time worker with an associate’s degree by $30 thousand. Combined, three full-time workers with each of these educational attainments earn $195 thousand before taxes. Find the median yearly before-tax income of full-time workers with each of these levels of education. (These incomes are illustrated by the bar graph on the previous page.)

Example 2 Modeling Attitudes of College Freshmen

Your author teaching math in 1969

Researchers have surveyed college freshmen every year since 1969. Figure P.15 shows that attitudes about some life goals have changed dramatically over the years. In particular, the freshman class of 2018 was more interested in making money than the freshmen of 1969 had been. In 1969, 42% of first-year college students considered “being well-off financially” essential or very important. For the period from 1969 through 2018, this percentage increased by approximately 0.8 each year. If this trend continues, by which year will all college freshmen consider “being well-off financially” essential or very important?

Figure P.15

A bar chart shows the life objectives of college freshmen, for the period from 19 69 to 20 18.

Source: Higher Education Research Institute

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We are interested in the year when all college freshmen, or 100% of the freshmen, will consider this life objective essential or very important. Let

    x = the number of years after 1969 when all freshmen will considerbeing    well-off financially essential or very important.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. There are no other unknown quantities to find, so we can skip this step.

  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS.

    The image shows a mathematical expression 42 + 0.8 x = 100
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    42+0.8x=100This is the equation that models the problems conditions.4242+0.8x=10042Subtract 42 from both sides.0.8x=58Simplify.0.8x0.8=580.8Divide both sides by 0.8.x=72.573Simplify and round to the nearestwhole number.

    Using current trends, by approximately 73 years after 1969, or in 2042, all freshmen will consider “being well-off financially” essential or very important.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that all freshmen (100%, represented by 100 using the model) will consider the objective essential or very important. Does this approximately occur if we increase the 1969 percentage, 42, by 0.8 each year for 73 years, our proposed solution?

    42+0.8(73)=42+58.4=100.4100

    This verifies that using trends shown in Figure P.15, all first-year college students will consider the objective of being well-off financially essential or very important approximately 73 years after 1969, or in 2042.

Check Point 2

  • Figure P.15 on the previous page shows that the freshman class of 2018 was less interested in developing a philosophy of life than the freshmen of 1969 had been. In 1969, 85% of the freshmen considered this objective essential or very important. Since then, this percentage has decreased by approximately 0.8 each year. If this trend continues, by which year will only 25% of college freshmen consider “developing a meaningful philosophy of life” essential or very important?

Example 3 Modeling Options for a Toll

It’s time to decide how you’re going to pay the bridge toll so that you can get to the beach. The first option requires purchasing a transponder for $20; with the transponder, you pay a reduced toll of $3.25 each time you cross the bridge. Your second option is toll-by-plate; with this option, you pay the full toll of $4.25 each time you cross the bridge plus a $3 administrative fee for each crossing. Find the number of times you would need to cross the bridge for the costs of the two options to be the same.

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. Let

    x=the number of times you cross the bridge.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. There are no other unknown quantities, so we can skip this step.

  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. The cost with the transponder is the cost of the transponder, $20, plus the toll, $3.25, times the number of crossings, x. For toll-by-plate, you must pay both the toll, $4.25, and the administrative fee, $3, each time you cross the bridge. The cost for toll-by-plate is the sum of the toll and the administrative fee, $4.25+$3 or $7.25, times the number of crossings, x. We want to know when these two costs are equal.

    The image shows a mathematical expression 7.25x = 20 + 3.25 x
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    7.25x=20+3.25xThis is the equation that models theproblems conditions.4x=20Subtract 3.25x from both sides.4x=5Divide both sides by 4.

    Because x represents the number of times you cross the bridge, the total cost of toll-by-plate is the same as the total cost with the transponder for 5 crossings.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that the cost of toll-by-plate should be the same as the cost with the transponder. Let’s see if they are the same with 5 bridge crossings per month.

    Expression to check the proposed solution.

    If you cross the bridge five times, both options cost $36.25. Thus the proposed solution, 5 crossings, satisfies the problem’s conditions.

Check Point 3

  • You drive up to a toll plaza and find booths with attendants where you can pay the toll by cash or credit card. With this option, the toll is $5 each time you cross the bridge. The attendant gives you the option of buying a bar-coded decal for $25; with the decal, you get 25% off the normal toll of $5 for each crossing. Find the number of times you would need to cross the bridge for the costs of the two options to be the same.

Example 4 A Price Reduction on Wireless Headphones

Your favorite electronics retailer is having a terrific sale on noise cancelling wireless headphones. (You’ll finally be able to listen to music in peace!) After a 40% price reduction, you purchase headphones for $192. What was the price of the headphones before the reduction?

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the price of the headphones prior to the reduction.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. There are no other unknown quantities to find, so we can skip this step.

  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. The headphones’ original price minus the 40% reduction is the reduced price, $192.

    The image shows a mathematical expression x minus 0.4 x = 192.
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    The solution of an equation.

    The headphones’ price before the reduction was $320.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The price before the reduction, $320, minus the 40% reduction should equal the reduced price given in the original wording, $192:

    32040% of 320=3200.4(320)=320128=192.

    This verifies that the headphones’ price before the reduction was $320.

Check Point 4

  • After a 30% price reduction, you purchase a new laptop for $840. What was the laptop’s price before the reduction?

Our next example is about simple interest. Simple interest involves interest calculated only on the amount of money that we invest, called the principal. The formula I=Pr is used to find the simple interest, I, earned for one year when the principal, P, is invested at an annual interest rate, r. Dual investment problems involve different amounts of money in two or more investments, each paying a different rate.

Example 5 Solving a Dual Investment Problem

Your grandmother needs your help. She has $150,000 to invest. Part of this money is to be invested in a money market account paying 1.5% annual interest. The rest of this money is to be invested in a certificate of deposit paying 1.3% annual interest. She told you that she requires $2000 per year in extra income from the combination of these investments. How much money should be placed in each investment?

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the amount invested in the money market account at 1.5%.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. The other quantity that we seek is the amount invested at 1.3% in the certificate of deposit. Because the total amount Grandma has to invest is $150,000 and we already used up x,

    150,000x=the amount invested in the certificate of deposit at 1.3%.
  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. Because Grandma requires $2000 in total interest, the interest for the two investments combined must be $2000. Interest is Pr or rP for each investment.

    The image shows a mathematical expression 0.015 x + 0.013 left parenthesis 150,000 minus x right parenthesis = 20 00
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    A mathematical expression steps to solve an equation to find the value of x.

    Thus,

    the amount invested at1.5%=x=25,000the amount invested at 1.3%=150,00025,000=125,000

    Grandma should invest $25,000 at 1.5% and $125,000 at 1.3%.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that the total interest from the dual investments should be $2000. Can Grandma count on $2000 interest? The interest earned on $25,000 at 1.5% is ($25,000) (0.015), or $375. The interest earned on $125,000 at 1.3% is ($125,000)(0.013), or $1625. The total interest is $375+$1625, or $2000, exactly as it should be. You’ve made your grandmother happy. (Now if you would just visit her more often …)

Check Point 5

  • You inherited $50,000 with the stipulation that for the first year the money had to be invested in two accounts paying 0.9% and 1.1% annual interest. How much did you invest at each rate if the total interest earned for the year was $515?

Solving geometry problems usually requires a knowledge of basic geometric ideas and formulas. Formulas for area, perimeter, and volume are given in Table P.6.

Table P.6 Common Formulas for Area, Perimeter, and Volume

An illustration shows common formulas for area, perimeter, and volume of different shapes.
Table P.6 Full Alternative Text

We will be using the formula for the perimeter of a rectangle, P=2l+2w, in our next example. The formula states that a rectangle’s perimeter is the sum of twice its length and twice its width.

Example 6 Finding the Dimensions of an American Football Field

The length of an American football field is 200 feet more than the width. If the perimeter of the field is 1040 feet, what are its dimensions?

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We know something about the length; the length is 200 feet more than the width. We will let

    x=the width.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. Because the length is 200 feet more than the width, we add 200 to the width to represent the length. Thus,

    x+200=the length.

    Figure P.16 illustrates an American football field and its dimensions.

    Figure P.16 An American football field

    An illustration shows an American football field with its width as x and length as x + 20 0.
  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. Because the perimeter of the field is 1040 feet,

    The image shows the mathematical expression 2 left parenthesis x + 20 0 right parenthesis + 2 x = 1040.

  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    Steps to solve an equation.

    Thus,

    width=x=160.length=x+200=160+200=360.

    The dimensions of an American football field are 160 feet by 360 feet. (The 360-foot length is usually described as 120 yards.)

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The perimeter of the football field using the dimensions that we found is

    2(160 feet)+2(360 feet)=320 feet+720 feet=1040 feet.

    Because the problem’s wording tells us that the perimeter is 1040 feet, our dimensions are correct.

Check Point 6

  • The length of a rectangular basketball court is 44 feet more than the width. If the perimeter of the basketball court is 288 feet, what are its dimensions?

We will use the formula for the area of a rectangle, A=lw, in our next example. The formula states that a rectangle’s area is the product of its length and its width.

Example 7 Solving a Problem Involving Landscape Design

A rectangular garden measures 80 feet by 60 feet. A large path of uniform width is to be added along both shorter sides and one longer side of the garden. The landscape designer doing the work wants to double the garden’s area with the addition of this path. How wide should the path be?

Solution

  1. Step 1 LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the width of the path.

    The situation is illustrated in Figure P.17. The figure shows the original 80-by-60 foot rectangular garden and the path of width x added along both shorter sides and one longer side.

    Figure P.17 The garden’s area is to be doubled by adding the path.

    An image shows a rectangular garden is surrounded by a path that has a length of 80 + 2 x and a width of 60 + x.
  2. Step 2 REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. Because the path is added along both shorter sides and one longer side, Figure P.17 shows that

     80+2x=the length of the new, expanded rectangle 60+x=the width of the new, expanded rectangle.

  3. Step 3 WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. The area of the rectangle must be doubled by the addition of the path.

    Left parenthesis 80 + 2 x right parenthesis, left parenthesis 60 + x right parenthesis = 2 times 80 times 60.
  4. Step 4 SOLVE THE EQUATION AND ANSWER THE QUESTION.

    (80+2x)(60+x)=28060This is the equation that modelsthe problems conditions.4800+200x+2x2=9600Multiply. Use FOIL on the left side.2x2+200x4800=0Subtract 9600 from both sidesand write the quadratic equationin standard form.2(x2+100x2400)=0Factor out 2, the GCF.2(x20)(x+120)=0Factor the trinomial.x20=0or x+120=0Set each variable factor equal to 0.x=20x=120Solve for x.

    The path cannot have a negative width. Because 120 is geometrically impossible, we use x=20. The width of the path should be 20 feet.

  5. Step 5 CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. Has the landscape architect doubled the garden’s area with the 20-foot-wide path? The area of the garden is 80 feet times 60 feet, or 4800 square feet. Because 80+2x and 60+x represent the length and width of the expanded rectangle,

    80+2x=80+220 =120 feet is the expanded rectangle's length.60+2x=60+20=80 feet is the expanded rectangle's width.

    The area of the expanded rectangle is 120 feet times 80 feet, or 9600 square feet. This is double the area of the garden, 4800 square feet, as specified by the problem’s conditions.

Check Point 7

  • A rectangular garden measures 16 feet by 12 feet. A path of uniform width is to be added so as to surround the entire garden. The landscape artist doing the work wants the garden and path to cover an area of 320 square feet. How wide should the path be?

In our next example, we will be using the Pythagorean Theorem to obtain a mathematical model. The ancient Greek philosopher and mathematician Pythagoras (approximately 582–500 b.c.) founded a school whose motto was “All is number.” Pythagoras is best remembered for his work with the right triangle, a triangle with one angle measuring 90°. The side opposite the 90° angle is called the hypotenuse. The other sides are called legs. Pythagoras found that if he constructed squares on each of the legs, as well as a larger square on the hypotenuse, the sum of the areas of the smaller squares is equal to the area of the larger square. This is illustrated in Figure P.18.

Figure P.18The area of the large square equals the sum of the areas of the smaller squares.

An illustration shows a right triangle of base 3, height 4, and hypotenuse 5.
Figure P. 18 Full Alternative Text

This relationship is usually stated in terms of the lengths of the three sides of a right triangle and is called the Pythagorean Theorem.

The Pythagorean Theorem

The sum of the squares of the lengths of the legs of a right triangle equals the square of the length of the hypotenuse.

If the legs have lengths a and b, and the hypotenuse has length c, then

a2+b2=c2.
An image shows a right angle triangle ABC, which represents angle C is a 90 degree angle. AC represents as a base labeled, b, Leg, BC denotes as a height labeled, a Leg, and AB is labeled, c, hypotenuse.

Example 8 Using the Pythagorean Theorem: Screen Math

A bar graph shows the rising percentage of large screen TVs on store shelves.

Source: GAP Intelligence

Did you know that the size of a television screen refers to the length of its diagonal? If the length of the HDTV screen in Figure P.19 is 28 inches and its width is 15.7 inches, what is the size of the screen to the nearest inch?

Figure P.19

A photo shows a television with a screen size of length 28 inches and a width of 15.7 inches.

Solution

Figure P.19 shows that the length, width, and diagonal of the screen form a right triangle. The diagonal is the hypotenuse of the triangle. We use the Pythagorean Theorem with a=28, b=15.7, and solve for the screen size, c.

a2+b2=c2This is the Pythagorean Theorem.282+15.72=c2Let a=28 and b=15.7.784+246.49=c2282=2828=784 and15.72=15.715.7=246.49.c2=1030.49Add: 784+246.49=1030.49. We also reversedthe two sides.
c=1030.49 or  c=1030.49Apply the square root property.c32 or  c32Use a calculator and round to thenearest inch. 1030.49    = or   1030.49 ENTER

Because c represents the size of the screen, this dimension must be positive. We reject 32. Thus, the screen size of the HDTV is 32 inches.

Check Point 8

  • Figure P.20 shows the dimensions of an old TV screen. What is the size of the screen?

    Figure P.20

    A photo shows a television with a screen size of length 25.6 inches and a width of 19.2 inches.

Example 9 Dividing the Cost of a Yacht

A group of friends agrees to share the cost of a $50,000 yacht equally. Before the purchase is made, one more person joins the group. As a result, each person’s share is reduced by $2500. How many people were in the original group?

Solution

  1. Step 1 LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the number of people in the original group.
  2. Step 2 REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x.  Because one more person joined the original group, let

    x+1=the number of people in the final group.
  3. Step 3 WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. As a result of one more person joining the original group, each person’s share is reduced by $2500.

    The image shows a mathematical expression, start fraction 50,000 over x end fraction minus 2500 = start fraction 50,000 over x + 1 end fraction.
  4. Step 4 SOLVE THE EQUATION AND ANSWER THE QUESTION.

    50,000x2500=50,000x+1This is the equation that models theproblems conditions.x(x+1)(50,000x2500)=x(x+1)50,000x+1Multiply both sides by x(x+1),the LCD. x (x+1)50,000 x x(x+1)2500=x (x+1) 50,000 (x+1) Use the distributive property anddivide out common factors.50,000(x+1)2500x(x+1)=50,000xSimplify.50,000x+50,0002500x22500x=50,000xUse the distributive property.2500x2+47,500x+50,000=50,000xCombine like terms.2500x22500x+50,000=0Write the quadratic equation instandard form.2500(x2+x20)=0Factor out 2500.2500(x+5)(x4)=0Factor completely.x+5=0or x4=0Set each variable factor equal to zero.x=5x=4Solve the resulting equations.

    Because x represents the number of people in the original group, x cannot be negative. Thus, there were four people in the original group.

  5. Step 5 CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM.

     original cost per person=$50,0004=$12,500 final cost per person=$50,0005=$10,000

    We see that each person’s share is reduced by $12,500$10,000, or $2500, as specified by the problem’s conditions.

Check Point 9

  • A group of people share equally in a $5,000,000 lottery. Before the money is divided, three more winning ticket holders are declared. As a result, each person’s share is reduced by $375,000. How many people were in the original group of winners?

Objective 1: Use equations to solve problems

Problem Solving with Equations

  1. Objective 1Use equations to solve problems.

Watch Video

We have seen that a model is a mathematical representation of a real-world situation. In this section, we will be solving problems that are presented in English. This means that we must obtain models by translating from the ordinary language of English into the language of algebraic equations. To translate, however, we must understand the English prose and be familiar with the forms of algebraic language. Following are some general steps we will use in solving word problems:

Strategy for Solving Word Problems

  1. Step 1 Read the problem carefully several times until you can state in your own words what is given and what the problem is looking for. Let x (or any variable) represent one of the unknown quantities in the problem.

  2. Step 2 If necessary, write expressions for any other unknown quantities in the problem in terms of x.

  3. Step 3 Write an equation in x that models the verbal conditions of the problem.

  4. Step 4 Solve the equation and answer the problem’s question.

  5. Step 5 Check the solution in the original wording of the problem, not in the equation obtained from the words.

Example 1 Education Pays Off

The graph in Figure P.14 shows median yearly earnings in the United States by highest educational attainment. It certainly looks like you can make more money with more education. However, workers who make more money pay a higher percentage of income in taxes, so it is wise to consider after-tax earnings when deciding if that higher degree will pay off.

Figure P.14

A bar chart shows median earnings of full-time workers in the United States, by highest educational attainment.

Source: Education Pays 2019

The median yearly after-tax income of a full-time worker with a bachelor’s degree exceeds that of a full-time worker with an associate’s degree by $11 thousand. The median yearly after-tax income of a full-time worker with a master’s degree exceeds that of a full-time worker with an associate’s degree by $21 thousand. Combined, three full-time workers with each of these degrees earn $149 thousand after taxes. Find the median yearly after-tax income of full-time workers with each of these levels of education.

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We know something about after-tax incomes of full-time workers with bachelor’s degrees and master’s degrees: They exceed the after-tax income of a full-time worker with an associate’s degree by $11 thousand and $21 thousand, respectively. We will let

    x=the median yearly after-tax income of a full-time worker with an associates degree.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. Because a full-time worker with a bachelor’s degree earns $11 thousand more after taxes than a full-time worker with an associate’s degree, let

    x+11=the median yearly after-tax income of a full-time worker with a bachelors degree.

    Because a full-time worker with a master’s degree earns $21 thousand more after taxes than a full-time worker with an associate’s degree, let

    x+21=the median yearly after-tax income of a full-time worker with a master's degree.
  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. Combined, three full-time workers with each of these degrees earn $149 thousand.

    An equation with variable x to model the conditions.

  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    x+(x+11)+(x+21)=149This is the equation that models the problems conditions.3x+32=149Remove parentheses, regroup, and combine like terms.3x=117Subtract 32 from both sides.x=39Divide both sides by 3.

    Because we isolated the variable in the model and obtained x=39,

    median after-tax income with an associates degree=x=39median after-tax income with a bachelors degree=x+11=39+11=50median after-tax income with a masters degree=x+21=39+21=60.

    Full-time workers with associate’s degrees earn $39 thousand per year after taxes, full-time workers with bachelor’s degrees earn $50 thousand per year after taxes, and full-time workers with master’s degrees earn $60 thousand per year after taxes.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that combined, three full-time workers with each of these educational attainments earn $149 thousand after taxes. Using the incomes we determined in step 4, the sum is

    $39 thousand+$50 thousand+$60 thousand,or $149 thousand,

    which satisfies the problem’s conditions.

Check Point 1

  • The median yearly before-tax income of a full-time worker with a bachelor’s degree exceeds that of a full-time worker with an associate’s degree by $15 thousand. The median yearly before-tax income of a full-time worker with a master’s degree exceeds that of a full-time worker with an associate’s degree by $30 thousand. Combined, three full-time workers with each of these educational attainments earn $195 thousand before taxes. Find the median yearly before-tax income of full-time workers with each of these levels of education. (These incomes are illustrated by the bar graph on the previous page.)

Example 2 Modeling Attitudes of College Freshmen

Your author teaching math in 1969

Researchers have surveyed college freshmen every year since 1969. Figure P.15 shows that attitudes about some life goals have changed dramatically over the years. In particular, the freshman class of 2018 was more interested in making money than the freshmen of 1969 had been. In 1969, 42% of first-year college students considered “being well-off financially” essential or very important. For the period from 1969 through 2018, this percentage increased by approximately 0.8 each year. If this trend continues, by which year will all college freshmen consider “being well-off financially” essential or very important?

Figure P.15

A bar chart shows the life objectives of college freshmen, for the period from 19 69 to 20 18.

Source: Higher Education Research Institute

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We are interested in the year when all college freshmen, or 100% of the freshmen, will consider this life objective essential or very important. Let

    x = the number of years after 1969 when all freshmen will considerbeing    well-off financially essential or very important.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. There are no other unknown quantities to find, so we can skip this step.

  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS.

    The image shows a mathematical expression 42 + 0.8 x = 100
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    42+0.8x=100This is the equation that models the problems conditions.4242+0.8x=10042Subtract 42 from both sides.0.8x=58Simplify.0.8x0.8=580.8Divide both sides by 0.8.x=72.573Simplify and round to the nearestwhole number.

    Using current trends, by approximately 73 years after 1969, or in 2042, all freshmen will consider “being well-off financially” essential or very important.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that all freshmen (100%, represented by 100 using the model) will consider the objective essential or very important. Does this approximately occur if we increase the 1969 percentage, 42, by 0.8 each year for 73 years, our proposed solution?

    42+0.8(73)=42+58.4=100.4100

    This verifies that using trends shown in Figure P.15, all first-year college students will consider the objective of being well-off financially essential or very important approximately 73 years after 1969, or in 2042.

Check Point 2

  • Figure P.15 on the previous page shows that the freshman class of 2018 was less interested in developing a philosophy of life than the freshmen of 1969 had been. In 1969, 85% of the freshmen considered this objective essential or very important. Since then, this percentage has decreased by approximately 0.8 each year. If this trend continues, by which year will only 25% of college freshmen consider “developing a meaningful philosophy of life” essential or very important?

Example 3 Modeling Options for a Toll

It’s time to decide how you’re going to pay the bridge toll so that you can get to the beach. The first option requires purchasing a transponder for $20; with the transponder, you pay a reduced toll of $3.25 each time you cross the bridge. Your second option is toll-by-plate; with this option, you pay the full toll of $4.25 each time you cross the bridge plus a $3 administrative fee for each crossing. Find the number of times you would need to cross the bridge for the costs of the two options to be the same.

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. Let

    x=the number of times you cross the bridge.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. There are no other unknown quantities, so we can skip this step.

  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. The cost with the transponder is the cost of the transponder, $20, plus the toll, $3.25, times the number of crossings, x. For toll-by-plate, you must pay both the toll, $4.25, and the administrative fee, $3, each time you cross the bridge. The cost for toll-by-plate is the sum of the toll and the administrative fee, $4.25+$3 or $7.25, times the number of crossings, x. We want to know when these two costs are equal.

    The image shows a mathematical expression 7.25x = 20 + 3.25 x
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    7.25x=20+3.25xThis is the equation that models theproblems conditions.4x=20Subtract 3.25x from both sides.4x=5Divide both sides by 4.

    Because x represents the number of times you cross the bridge, the total cost of toll-by-plate is the same as the total cost with the transponder for 5 crossings.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that the cost of toll-by-plate should be the same as the cost with the transponder. Let’s see if they are the same with 5 bridge crossings per month.

    Expression to check the proposed solution.

    If you cross the bridge five times, both options cost $36.25. Thus the proposed solution, 5 crossings, satisfies the problem’s conditions.

Check Point 3

  • You drive up to a toll plaza and find booths with attendants where you can pay the toll by cash or credit card. With this option, the toll is $5 each time you cross the bridge. The attendant gives you the option of buying a bar-coded decal for $25; with the decal, you get 25% off the normal toll of $5 for each crossing. Find the number of times you would need to cross the bridge for the costs of the two options to be the same.

Example 4 A Price Reduction on Wireless Headphones

Your favorite electronics retailer is having a terrific sale on noise cancelling wireless headphones. (You’ll finally be able to listen to music in peace!) After a 40% price reduction, you purchase headphones for $192. What was the price of the headphones before the reduction?

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the price of the headphones prior to the reduction.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. There are no other unknown quantities to find, so we can skip this step.

  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. The headphones’ original price minus the 40% reduction is the reduced price, $192.

    The image shows a mathematical expression x minus 0.4 x = 192.
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    The solution of an equation.

    The headphones’ price before the reduction was $320.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The price before the reduction, $320, minus the 40% reduction should equal the reduced price given in the original wording, $192:

    32040% of 320=3200.4(320)=320128=192.

    This verifies that the headphones’ price before the reduction was $320.

Check Point 4

  • After a 30% price reduction, you purchase a new laptop for $840. What was the laptop’s price before the reduction?

Our next example is about simple interest. Simple interest involves interest calculated only on the amount of money that we invest, called the principal. The formula I=Pr is used to find the simple interest, I, earned for one year when the principal, P, is invested at an annual interest rate, r. Dual investment problems involve different amounts of money in two or more investments, each paying a different rate.

Example 5 Solving a Dual Investment Problem

Your grandmother needs your help. She has $150,000 to invest. Part of this money is to be invested in a money market account paying 1.5% annual interest. The rest of this money is to be invested in a certificate of deposit paying 1.3% annual interest. She told you that she requires $2000 per year in extra income from the combination of these investments. How much money should be placed in each investment?

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the amount invested in the money market account at 1.5%.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. The other quantity that we seek is the amount invested at 1.3% in the certificate of deposit. Because the total amount Grandma has to invest is $150,000 and we already used up x,

    150,000x=the amount invested in the certificate of deposit at 1.3%.
  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. Because Grandma requires $2000 in total interest, the interest for the two investments combined must be $2000. Interest is Pr or rP for each investment.

    The image shows a mathematical expression 0.015 x + 0.013 left parenthesis 150,000 minus x right parenthesis = 20 00
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    A mathematical expression steps to solve an equation to find the value of x.

    Thus,

    the amount invested at1.5%=x=25,000the amount invested at 1.3%=150,00025,000=125,000

    Grandma should invest $25,000 at 1.5% and $125,000 at 1.3%.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that the total interest from the dual investments should be $2000. Can Grandma count on $2000 interest? The interest earned on $25,000 at 1.5% is ($25,000) (0.015), or $375. The interest earned on $125,000 at 1.3% is ($125,000)(0.013), or $1625. The total interest is $375+$1625, or $2000, exactly as it should be. You’ve made your grandmother happy. (Now if you would just visit her more often …)

Check Point 5

  • You inherited $50,000 with the stipulation that for the first year the money had to be invested in two accounts paying 0.9% and 1.1% annual interest. How much did you invest at each rate if the total interest earned for the year was $515?

Solving geometry problems usually requires a knowledge of basic geometric ideas and formulas. Formulas for area, perimeter, and volume are given in Table P.6.

Table P.6 Common Formulas for Area, Perimeter, and Volume

An illustration shows common formulas for area, perimeter, and volume of different shapes.
Table P.6 Full Alternative Text

We will be using the formula for the perimeter of a rectangle, P=2l+2w, in our next example. The formula states that a rectangle’s perimeter is the sum of twice its length and twice its width.

Example 6 Finding the Dimensions of an American Football Field

The length of an American football field is 200 feet more than the width. If the perimeter of the field is 1040 feet, what are its dimensions?

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We know something about the length; the length is 200 feet more than the width. We will let

    x=the width.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. Because the length is 200 feet more than the width, we add 200 to the width to represent the length. Thus,

    x+200=the length.

    Figure P.16 illustrates an American football field and its dimensions.

    Figure P.16 An American football field

    An illustration shows an American football field with its width as x and length as x + 20 0.
  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. Because the perimeter of the field is 1040 feet,

    The image shows the mathematical expression 2 left parenthesis x + 20 0 right parenthesis + 2 x = 1040.

  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    Steps to solve an equation.

    Thus,

    width=x=160.length=x+200=160+200=360.

    The dimensions of an American football field are 160 feet by 360 feet. (The 360-foot length is usually described as 120 yards.)

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The perimeter of the football field using the dimensions that we found is

    2(160 feet)+2(360 feet)=320 feet+720 feet=1040 feet.

    Because the problem’s wording tells us that the perimeter is 1040 feet, our dimensions are correct.

Check Point 6

  • The length of a rectangular basketball court is 44 feet more than the width. If the perimeter of the basketball court is 288 feet, what are its dimensions?

We will use the formula for the area of a rectangle, A=lw, in our next example. The formula states that a rectangle’s area is the product of its length and its width.

Example 7 Solving a Problem Involving Landscape Design

A rectangular garden measures 80 feet by 60 feet. A large path of uniform width is to be added along both shorter sides and one longer side of the garden. The landscape designer doing the work wants to double the garden’s area with the addition of this path. How wide should the path be?

Solution

  1. Step 1 LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the width of the path.

    The situation is illustrated in Figure P.17. The figure shows the original 80-by-60 foot rectangular garden and the path of width x added along both shorter sides and one longer side.

    Figure P.17 The garden’s area is to be doubled by adding the path.

    An image shows a rectangular garden is surrounded by a path that has a length of 80 + 2 x and a width of 60 + x.
  2. Step 2 REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. Because the path is added along both shorter sides and one longer side, Figure P.17 shows that

     80+2x=the length of the new, expanded rectangle 60+x=the width of the new, expanded rectangle.

  3. Step 3 WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. The area of the rectangle must be doubled by the addition of the path.

    Left parenthesis 80 + 2 x right parenthesis, left parenthesis 60 + x right parenthesis = 2 times 80 times 60.
  4. Step 4 SOLVE THE EQUATION AND ANSWER THE QUESTION.

    (80+2x)(60+x)=28060This is the equation that modelsthe problems conditions.4800+200x+2x2=9600Multiply. Use FOIL on the left side.2x2+200x4800=0Subtract 9600 from both sidesand write the quadratic equationin standard form.2(x2+100x2400)=0Factor out 2, the GCF.2(x20)(x+120)=0Factor the trinomial.x20=0or x+120=0Set each variable factor equal to 0.x=20x=120Solve for x.

    The path cannot have a negative width. Because 120 is geometrically impossible, we use x=20. The width of the path should be 20 feet.

  5. Step 5 CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. Has the landscape architect doubled the garden’s area with the 20-foot-wide path? The area of the garden is 80 feet times 60 feet, or 4800 square feet. Because 80+2x and 60+x represent the length and width of the expanded rectangle,

    80+2x=80+220 =120 feet is the expanded rectangle's length.60+2x=60+20=80 feet is the expanded rectangle's width.

    The area of the expanded rectangle is 120 feet times 80 feet, or 9600 square feet. This is double the area of the garden, 4800 square feet, as specified by the problem’s conditions.

Check Point 7

  • A rectangular garden measures 16 feet by 12 feet. A path of uniform width is to be added so as to surround the entire garden. The landscape artist doing the work wants the garden and path to cover an area of 320 square feet. How wide should the path be?

In our next example, we will be using the Pythagorean Theorem to obtain a mathematical model. The ancient Greek philosopher and mathematician Pythagoras (approximately 582–500 b.c.) founded a school whose motto was “All is number.” Pythagoras is best remembered for his work with the right triangle, a triangle with one angle measuring 90°. The side opposite the 90° angle is called the hypotenuse. The other sides are called legs. Pythagoras found that if he constructed squares on each of the legs, as well as a larger square on the hypotenuse, the sum of the areas of the smaller squares is equal to the area of the larger square. This is illustrated in Figure P.18.

Figure P.18The area of the large square equals the sum of the areas of the smaller squares.

An illustration shows a right triangle of base 3, height 4, and hypotenuse 5.
Figure P. 18 Full Alternative Text

This relationship is usually stated in terms of the lengths of the three sides of a right triangle and is called the Pythagorean Theorem.

The Pythagorean Theorem

The sum of the squares of the lengths of the legs of a right triangle equals the square of the length of the hypotenuse.

If the legs have lengths a and b, and the hypotenuse has length c, then

a2+b2=c2.
An image shows a right angle triangle ABC, which represents angle C is a 90 degree angle. AC represents as a base labeled, b, Leg, BC denotes as a height labeled, a Leg, and AB is labeled, c, hypotenuse.

Example 8 Using the Pythagorean Theorem: Screen Math

A bar graph shows the rising percentage of large screen TVs on store shelves.

Source: GAP Intelligence

Did you know that the size of a television screen refers to the length of its diagonal? If the length of the HDTV screen in Figure P.19 is 28 inches and its width is 15.7 inches, what is the size of the screen to the nearest inch?

Figure P.19

A photo shows a television with a screen size of length 28 inches and a width of 15.7 inches.

Solution

Figure P.19 shows that the length, width, and diagonal of the screen form a right triangle. The diagonal is the hypotenuse of the triangle. We use the Pythagorean Theorem with a=28, b=15.7, and solve for the screen size, c.

a2+b2=c2This is the Pythagorean Theorem.282+15.72=c2Let a=28 and b=15.7.784+246.49=c2282=2828=784 and15.72=15.715.7=246.49.c2=1030.49Add: 784+246.49=1030.49. We also reversedthe two sides.
c=1030.49 or  c=1030.49Apply the square root property.c32 or  c32Use a calculator and round to thenearest inch. 1030.49    = or   1030.49 ENTER

Because c represents the size of the screen, this dimension must be positive. We reject 32. Thus, the screen size of the HDTV is 32 inches.

Check Point 8

  • Figure P.20 shows the dimensions of an old TV screen. What is the size of the screen?

    Figure P.20

    A photo shows a television with a screen size of length 25.6 inches and a width of 19.2 inches.

Example 9 Dividing the Cost of a Yacht

A group of friends agrees to share the cost of a $50,000 yacht equally. Before the purchase is made, one more person joins the group. As a result, each person’s share is reduced by $2500. How many people were in the original group?

Solution

  1. Step 1 LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the number of people in the original group.
  2. Step 2 REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x.  Because one more person joined the original group, let

    x+1=the number of people in the final group.
  3. Step 3 WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. As a result of one more person joining the original group, each person’s share is reduced by $2500.

    The image shows a mathematical expression, start fraction 50,000 over x end fraction minus 2500 = start fraction 50,000 over x + 1 end fraction.
  4. Step 4 SOLVE THE EQUATION AND ANSWER THE QUESTION.

    50,000x2500=50,000x+1This is the equation that models theproblems conditions.x(x+1)(50,000x2500)=x(x+1)50,000x+1Multiply both sides by x(x+1),the LCD. x (x+1)50,000 x x(x+1)2500=x (x+1) 50,000 (x+1) Use the distributive property anddivide out common factors.50,000(x+1)2500x(x+1)=50,000xSimplify.50,000x+50,0002500x22500x=50,000xUse the distributive property.2500x2+47,500x+50,000=50,000xCombine like terms.2500x22500x+50,000=0Write the quadratic equation instandard form.2500(x2+x20)=0Factor out 2500.2500(x+5)(x4)=0Factor completely.x+5=0or x4=0Set each variable factor equal to zero.x=5x=4Solve the resulting equations.

    Because x represents the number of people in the original group, x cannot be negative. Thus, there were four people in the original group.

  5. Step 5 CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM.

     original cost per person=$50,0004=$12,500 final cost per person=$50,0005=$10,000

    We see that each person’s share is reduced by $12,500$10,000, or $2500, as specified by the problem’s conditions.

Check Point 9

  • A group of people share equally in a $5,000,000 lottery. Before the money is divided, three more winning ticket holders are declared. As a result, each person’s share is reduced by $375,000. How many people were in the original group of winners?

Objective 1: Use equations to solve problems

Problem Solving with Equations

  1. Objective 1Use equations to solve problems.

Watch Video

We have seen that a model is a mathematical representation of a real-world situation. In this section, we will be solving problems that are presented in English. This means that we must obtain models by translating from the ordinary language of English into the language of algebraic equations. To translate, however, we must understand the English prose and be familiar with the forms of algebraic language. Following are some general steps we will use in solving word problems:

Strategy for Solving Word Problems

  1. Step 1 Read the problem carefully several times until you can state in your own words what is given and what the problem is looking for. Let x (or any variable) represent one of the unknown quantities in the problem.

  2. Step 2 If necessary, write expressions for any other unknown quantities in the problem in terms of x.

  3. Step 3 Write an equation in x that models the verbal conditions of the problem.

  4. Step 4 Solve the equation and answer the problem’s question.

  5. Step 5 Check the solution in the original wording of the problem, not in the equation obtained from the words.

Example 1 Education Pays Off

The graph in Figure P.14 shows median yearly earnings in the United States by highest educational attainment. It certainly looks like you can make more money with more education. However, workers who make more money pay a higher percentage of income in taxes, so it is wise to consider after-tax earnings when deciding if that higher degree will pay off.

Figure P.14

A bar chart shows median earnings of full-time workers in the United States, by highest educational attainment.

Source: Education Pays 2019

The median yearly after-tax income of a full-time worker with a bachelor’s degree exceeds that of a full-time worker with an associate’s degree by $11 thousand. The median yearly after-tax income of a full-time worker with a master’s degree exceeds that of a full-time worker with an associate’s degree by $21 thousand. Combined, three full-time workers with each of these degrees earn $149 thousand after taxes. Find the median yearly after-tax income of full-time workers with each of these levels of education.

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We know something about after-tax incomes of full-time workers with bachelor’s degrees and master’s degrees: They exceed the after-tax income of a full-time worker with an associate’s degree by $11 thousand and $21 thousand, respectively. We will let

    x=the median yearly after-tax income of a full-time worker with an associates degree.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. Because a full-time worker with a bachelor’s degree earns $11 thousand more after taxes than a full-time worker with an associate’s degree, let

    x+11=the median yearly after-tax income of a full-time worker with a bachelors degree.

    Because a full-time worker with a master’s degree earns $21 thousand more after taxes than a full-time worker with an associate’s degree, let

    x+21=the median yearly after-tax income of a full-time worker with a master's degree.
  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. Combined, three full-time workers with each of these degrees earn $149 thousand.

    An equation with variable x to model the conditions.

  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    x+(x+11)+(x+21)=149This is the equation that models the problems conditions.3x+32=149Remove parentheses, regroup, and combine like terms.3x=117Subtract 32 from both sides.x=39Divide both sides by 3.

    Because we isolated the variable in the model and obtained x=39,

    median after-tax income with an associates degree=x=39median after-tax income with a bachelors degree=x+11=39+11=50median after-tax income with a masters degree=x+21=39+21=60.

    Full-time workers with associate’s degrees earn $39 thousand per year after taxes, full-time workers with bachelor’s degrees earn $50 thousand per year after taxes, and full-time workers with master’s degrees earn $60 thousand per year after taxes.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that combined, three full-time workers with each of these educational attainments earn $149 thousand after taxes. Using the incomes we determined in step 4, the sum is

    $39 thousand+$50 thousand+$60 thousand,or $149 thousand,

    which satisfies the problem’s conditions.

Check Point 1

  • The median yearly before-tax income of a full-time worker with a bachelor’s degree exceeds that of a full-time worker with an associate’s degree by $15 thousand. The median yearly before-tax income of a full-time worker with a master’s degree exceeds that of a full-time worker with an associate’s degree by $30 thousand. Combined, three full-time workers with each of these educational attainments earn $195 thousand before taxes. Find the median yearly before-tax income of full-time workers with each of these levels of education. (These incomes are illustrated by the bar graph on the previous page.)

Example 2 Modeling Attitudes of College Freshmen

Your author teaching math in 1969

Researchers have surveyed college freshmen every year since 1969. Figure P.15 shows that attitudes about some life goals have changed dramatically over the years. In particular, the freshman class of 2018 was more interested in making money than the freshmen of 1969 had been. In 1969, 42% of first-year college students considered “being well-off financially” essential or very important. For the period from 1969 through 2018, this percentage increased by approximately 0.8 each year. If this trend continues, by which year will all college freshmen consider “being well-off financially” essential or very important?

Figure P.15

A bar chart shows the life objectives of college freshmen, for the period from 19 69 to 20 18.

Source: Higher Education Research Institute

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We are interested in the year when all college freshmen, or 100% of the freshmen, will consider this life objective essential or very important. Let

    x = the number of years after 1969 when all freshmen will considerbeing    well-off financially essential or very important.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. There are no other unknown quantities to find, so we can skip this step.

  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS.

    The image shows a mathematical expression 42 + 0.8 x = 100
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    42+0.8x=100This is the equation that models the problems conditions.4242+0.8x=10042Subtract 42 from both sides.0.8x=58Simplify.0.8x0.8=580.8Divide both sides by 0.8.x=72.573Simplify and round to the nearestwhole number.

    Using current trends, by approximately 73 years after 1969, or in 2042, all freshmen will consider “being well-off financially” essential or very important.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that all freshmen (100%, represented by 100 using the model) will consider the objective essential or very important. Does this approximately occur if we increase the 1969 percentage, 42, by 0.8 each year for 73 years, our proposed solution?

    42+0.8(73)=42+58.4=100.4100

    This verifies that using trends shown in Figure P.15, all first-year college students will consider the objective of being well-off financially essential or very important approximately 73 years after 1969, or in 2042.

Check Point 2

  • Figure P.15 on the previous page shows that the freshman class of 2018 was less interested in developing a philosophy of life than the freshmen of 1969 had been. In 1969, 85% of the freshmen considered this objective essential or very important. Since then, this percentage has decreased by approximately 0.8 each year. If this trend continues, by which year will only 25% of college freshmen consider “developing a meaningful philosophy of life” essential or very important?

Example 3 Modeling Options for a Toll

It’s time to decide how you’re going to pay the bridge toll so that you can get to the beach. The first option requires purchasing a transponder for $20; with the transponder, you pay a reduced toll of $3.25 each time you cross the bridge. Your second option is toll-by-plate; with this option, you pay the full toll of $4.25 each time you cross the bridge plus a $3 administrative fee for each crossing. Find the number of times you would need to cross the bridge for the costs of the two options to be the same.

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. Let

    x=the number of times you cross the bridge.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. There are no other unknown quantities, so we can skip this step.

  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. The cost with the transponder is the cost of the transponder, $20, plus the toll, $3.25, times the number of crossings, x. For toll-by-plate, you must pay both the toll, $4.25, and the administrative fee, $3, each time you cross the bridge. The cost for toll-by-plate is the sum of the toll and the administrative fee, $4.25+$3 or $7.25, times the number of crossings, x. We want to know when these two costs are equal.

    The image shows a mathematical expression 7.25x = 20 + 3.25 x
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    7.25x=20+3.25xThis is the equation that models theproblems conditions.4x=20Subtract 3.25x from both sides.4x=5Divide both sides by 4.

    Because x represents the number of times you cross the bridge, the total cost of toll-by-plate is the same as the total cost with the transponder for 5 crossings.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that the cost of toll-by-plate should be the same as the cost with the transponder. Let’s see if they are the same with 5 bridge crossings per month.

    Expression to check the proposed solution.

    If you cross the bridge five times, both options cost $36.25. Thus the proposed solution, 5 crossings, satisfies the problem’s conditions.

Check Point 3

  • You drive up to a toll plaza and find booths with attendants where you can pay the toll by cash or credit card. With this option, the toll is $5 each time you cross the bridge. The attendant gives you the option of buying a bar-coded decal for $25; with the decal, you get 25% off the normal toll of $5 for each crossing. Find the number of times you would need to cross the bridge for the costs of the two options to be the same.

Example 4 A Price Reduction on Wireless Headphones

Your favorite electronics retailer is having a terrific sale on noise cancelling wireless headphones. (You’ll finally be able to listen to music in peace!) After a 40% price reduction, you purchase headphones for $192. What was the price of the headphones before the reduction?

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the price of the headphones prior to the reduction.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. There are no other unknown quantities to find, so we can skip this step.

  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. The headphones’ original price minus the 40% reduction is the reduced price, $192.

    The image shows a mathematical expression x minus 0.4 x = 192.
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    The solution of an equation.

    The headphones’ price before the reduction was $320.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The price before the reduction, $320, minus the 40% reduction should equal the reduced price given in the original wording, $192:

    32040% of 320=3200.4(320)=320128=192.

    This verifies that the headphones’ price before the reduction was $320.

Check Point 4

  • After a 30% price reduction, you purchase a new laptop for $840. What was the laptop’s price before the reduction?

Our next example is about simple interest. Simple interest involves interest calculated only on the amount of money that we invest, called the principal. The formula I=Pr is used to find the simple interest, I, earned for one year when the principal, P, is invested at an annual interest rate, r. Dual investment problems involve different amounts of money in two or more investments, each paying a different rate.

Example 5 Solving a Dual Investment Problem

Your grandmother needs your help. She has $150,000 to invest. Part of this money is to be invested in a money market account paying 1.5% annual interest. The rest of this money is to be invested in a certificate of deposit paying 1.3% annual interest. She told you that she requires $2000 per year in extra income from the combination of these investments. How much money should be placed in each investment?

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the amount invested in the money market account at 1.5%.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. The other quantity that we seek is the amount invested at 1.3% in the certificate of deposit. Because the total amount Grandma has to invest is $150,000 and we already used up x,

    150,000x=the amount invested in the certificate of deposit at 1.3%.
  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. Because Grandma requires $2000 in total interest, the interest for the two investments combined must be $2000. Interest is Pr or rP for each investment.

    The image shows a mathematical expression 0.015 x + 0.013 left parenthesis 150,000 minus x right parenthesis = 20 00
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    A mathematical expression steps to solve an equation to find the value of x.

    Thus,

    the amount invested at1.5%=x=25,000the amount invested at 1.3%=150,00025,000=125,000

    Grandma should invest $25,000 at 1.5% and $125,000 at 1.3%.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that the total interest from the dual investments should be $2000. Can Grandma count on $2000 interest? The interest earned on $25,000 at 1.5% is ($25,000) (0.015), or $375. The interest earned on $125,000 at 1.3% is ($125,000)(0.013), or $1625. The total interest is $375+$1625, or $2000, exactly as it should be. You’ve made your grandmother happy. (Now if you would just visit her more often …)

Check Point 5

  • You inherited $50,000 with the stipulation that for the first year the money had to be invested in two accounts paying 0.9% and 1.1% annual interest. How much did you invest at each rate if the total interest earned for the year was $515?

Solving geometry problems usually requires a knowledge of basic geometric ideas and formulas. Formulas for area, perimeter, and volume are given in Table P.6.

Table P.6 Common Formulas for Area, Perimeter, and Volume

An illustration shows common formulas for area, perimeter, and volume of different shapes.
Table P.6 Full Alternative Text

We will be using the formula for the perimeter of a rectangle, P=2l+2w, in our next example. The formula states that a rectangle’s perimeter is the sum of twice its length and twice its width.

Example 6 Finding the Dimensions of an American Football Field

The length of an American football field is 200 feet more than the width. If the perimeter of the field is 1040 feet, what are its dimensions?

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We know something about the length; the length is 200 feet more than the width. We will let

    x=the width.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. Because the length is 200 feet more than the width, we add 200 to the width to represent the length. Thus,

    x+200=the length.

    Figure P.16 illustrates an American football field and its dimensions.

    Figure P.16 An American football field

    An illustration shows an American football field with its width as x and length as x + 20 0.
  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. Because the perimeter of the field is 1040 feet,

    The image shows the mathematical expression 2 left parenthesis x + 20 0 right parenthesis + 2 x = 1040.

  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    Steps to solve an equation.

    Thus,

    width=x=160.length=x+200=160+200=360.

    The dimensions of an American football field are 160 feet by 360 feet. (The 360-foot length is usually described as 120 yards.)

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The perimeter of the football field using the dimensions that we found is

    2(160 feet)+2(360 feet)=320 feet+720 feet=1040 feet.

    Because the problem’s wording tells us that the perimeter is 1040 feet, our dimensions are correct.

Check Point 6

  • The length of a rectangular basketball court is 44 feet more than the width. If the perimeter of the basketball court is 288 feet, what are its dimensions?

We will use the formula for the area of a rectangle, A=lw, in our next example. The formula states that a rectangle’s area is the product of its length and its width.

Example 7 Solving a Problem Involving Landscape Design

A rectangular garden measures 80 feet by 60 feet. A large path of uniform width is to be added along both shorter sides and one longer side of the garden. The landscape designer doing the work wants to double the garden’s area with the addition of this path. How wide should the path be?

Solution

  1. Step 1 LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the width of the path.

    The situation is illustrated in Figure P.17. The figure shows the original 80-by-60 foot rectangular garden and the path of width x added along both shorter sides and one longer side.

    Figure P.17 The garden’s area is to be doubled by adding the path.

    An image shows a rectangular garden is surrounded by a path that has a length of 80 + 2 x and a width of 60 + x.
  2. Step 2 REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. Because the path is added along both shorter sides and one longer side, Figure P.17 shows that

     80+2x=the length of the new, expanded rectangle 60+x=the width of the new, expanded rectangle.

  3. Step 3 WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. The area of the rectangle must be doubled by the addition of the path.

    Left parenthesis 80 + 2 x right parenthesis, left parenthesis 60 + x right parenthesis = 2 times 80 times 60.
  4. Step 4 SOLVE THE EQUATION AND ANSWER THE QUESTION.

    (80+2x)(60+x)=28060This is the equation that modelsthe problems conditions.4800+200x+2x2=9600Multiply. Use FOIL on the left side.2x2+200x4800=0Subtract 9600 from both sidesand write the quadratic equationin standard form.2(x2+100x2400)=0Factor out 2, the GCF.2(x20)(x+120)=0Factor the trinomial.x20=0or x+120=0Set each variable factor equal to 0.x=20x=120Solve for x.

    The path cannot have a negative width. Because 120 is geometrically impossible, we use x=20. The width of the path should be 20 feet.

  5. Step 5 CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. Has the landscape architect doubled the garden’s area with the 20-foot-wide path? The area of the garden is 80 feet times 60 feet, or 4800 square feet. Because 80+2x and 60+x represent the length and width of the expanded rectangle,

    80+2x=80+220 =120 feet is the expanded rectangle's length.60+2x=60+20=80 feet is the expanded rectangle's width.

    The area of the expanded rectangle is 120 feet times 80 feet, or 9600 square feet. This is double the area of the garden, 4800 square feet, as specified by the problem’s conditions.

Check Point 7

  • A rectangular garden measures 16 feet by 12 feet. A path of uniform width is to be added so as to surround the entire garden. The landscape artist doing the work wants the garden and path to cover an area of 320 square feet. How wide should the path be?

In our next example, we will be using the Pythagorean Theorem to obtain a mathematical model. The ancient Greek philosopher and mathematician Pythagoras (approximately 582–500 b.c.) founded a school whose motto was “All is number.” Pythagoras is best remembered for his work with the right triangle, a triangle with one angle measuring 90°. The side opposite the 90° angle is called the hypotenuse. The other sides are called legs. Pythagoras found that if he constructed squares on each of the legs, as well as a larger square on the hypotenuse, the sum of the areas of the smaller squares is equal to the area of the larger square. This is illustrated in Figure P.18.

Figure P.18The area of the large square equals the sum of the areas of the smaller squares.

An illustration shows a right triangle of base 3, height 4, and hypotenuse 5.
Figure P. 18 Full Alternative Text

This relationship is usually stated in terms of the lengths of the three sides of a right triangle and is called the Pythagorean Theorem.

The Pythagorean Theorem

The sum of the squares of the lengths of the legs of a right triangle equals the square of the length of the hypotenuse.

If the legs have lengths a and b, and the hypotenuse has length c, then

a2+b2=c2.
An image shows a right angle triangle ABC, which represents angle C is a 90 degree angle. AC represents as a base labeled, b, Leg, BC denotes as a height labeled, a Leg, and AB is labeled, c, hypotenuse.

Example 8 Using the Pythagorean Theorem: Screen Math

A bar graph shows the rising percentage of large screen TVs on store shelves.

Source: GAP Intelligence

Did you know that the size of a television screen refers to the length of its diagonal? If the length of the HDTV screen in Figure P.19 is 28 inches and its width is 15.7 inches, what is the size of the screen to the nearest inch?

Figure P.19

A photo shows a television with a screen size of length 28 inches and a width of 15.7 inches.

Solution

Figure P.19 shows that the length, width, and diagonal of the screen form a right triangle. The diagonal is the hypotenuse of the triangle. We use the Pythagorean Theorem with a=28, b=15.7, and solve for the screen size, c.

a2+b2=c2This is the Pythagorean Theorem.282+15.72=c2Let a=28 and b=15.7.784+246.49=c2282=2828=784 and15.72=15.715.7=246.49.c2=1030.49Add: 784+246.49=1030.49. We also reversedthe two sides.
c=1030.49 or  c=1030.49Apply the square root property.c32 or  c32Use a calculator and round to thenearest inch. 1030.49    = or   1030.49 ENTER

Because c represents the size of the screen, this dimension must be positive. We reject 32. Thus, the screen size of the HDTV is 32 inches.

Check Point 8

  • Figure P.20 shows the dimensions of an old TV screen. What is the size of the screen?

    Figure P.20

    A photo shows a television with a screen size of length 25.6 inches and a width of 19.2 inches.

Example 9 Dividing the Cost of a Yacht

A group of friends agrees to share the cost of a $50,000 yacht equally. Before the purchase is made, one more person joins the group. As a result, each person’s share is reduced by $2500. How many people were in the original group?

Solution

  1. Step 1 LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the number of people in the original group.
  2. Step 2 REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x.  Because one more person joined the original group, let

    x+1=the number of people in the final group.
  3. Step 3 WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. As a result of one more person joining the original group, each person’s share is reduced by $2500.

    The image shows a mathematical expression, start fraction 50,000 over x end fraction minus 2500 = start fraction 50,000 over x + 1 end fraction.
  4. Step 4 SOLVE THE EQUATION AND ANSWER THE QUESTION.

    50,000x2500=50,000x+1This is the equation that models theproblems conditions.x(x+1)(50,000x2500)=x(x+1)50,000x+1Multiply both sides by x(x+1),the LCD. x (x+1)50,000 x x(x+1)2500=x (x+1) 50,000 (x+1) Use the distributive property anddivide out common factors.50,000(x+1)2500x(x+1)=50,000xSimplify.50,000x+50,0002500x22500x=50,000xUse the distributive property.2500x2+47,500x+50,000=50,000xCombine like terms.2500x22500x+50,000=0Write the quadratic equation instandard form.2500(x2+x20)=0Factor out 2500.2500(x+5)(x4)=0Factor completely.x+5=0or x4=0Set each variable factor equal to zero.x=5x=4Solve the resulting equations.

    Because x represents the number of people in the original group, x cannot be negative. Thus, there were four people in the original group.

  5. Step 5 CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM.

     original cost per person=$50,0004=$12,500 final cost per person=$50,0005=$10,000

    We see that each person’s share is reduced by $12,500$10,000, or $2500, as specified by the problem’s conditions.

Check Point 9

  • A group of people share equally in a $5,000,000 lottery. Before the money is divided, three more winning ticket holders are declared. As a result, each person’s share is reduced by $375,000. How many people were in the original group of winners?

Objective 1: Use equations to solve problems

Problem Solving with Equations

  1. Objective 1Use equations to solve problems.

Watch Video

We have seen that a model is a mathematical representation of a real-world situation. In this section, we will be solving problems that are presented in English. This means that we must obtain models by translating from the ordinary language of English into the language of algebraic equations. To translate, however, we must understand the English prose and be familiar with the forms of algebraic language. Following are some general steps we will use in solving word problems:

Strategy for Solving Word Problems

  1. Step 1 Read the problem carefully several times until you can state in your own words what is given and what the problem is looking for. Let x (or any variable) represent one of the unknown quantities in the problem.

  2. Step 2 If necessary, write expressions for any other unknown quantities in the problem in terms of x.

  3. Step 3 Write an equation in x that models the verbal conditions of the problem.

  4. Step 4 Solve the equation and answer the problem’s question.

  5. Step 5 Check the solution in the original wording of the problem, not in the equation obtained from the words.

Example 1 Education Pays Off

The graph in Figure P.14 shows median yearly earnings in the United States by highest educational attainment. It certainly looks like you can make more money with more education. However, workers who make more money pay a higher percentage of income in taxes, so it is wise to consider after-tax earnings when deciding if that higher degree will pay off.

Figure P.14

A bar chart shows median earnings of full-time workers in the United States, by highest educational attainment.

Source: Education Pays 2019

The median yearly after-tax income of a full-time worker with a bachelor’s degree exceeds that of a full-time worker with an associate’s degree by $11 thousand. The median yearly after-tax income of a full-time worker with a master’s degree exceeds that of a full-time worker with an associate’s degree by $21 thousand. Combined, three full-time workers with each of these degrees earn $149 thousand after taxes. Find the median yearly after-tax income of full-time workers with each of these levels of education.

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We know something about after-tax incomes of full-time workers with bachelor’s degrees and master’s degrees: They exceed the after-tax income of a full-time worker with an associate’s degree by $11 thousand and $21 thousand, respectively. We will let

    x=the median yearly after-tax income of a full-time worker with an associates degree.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. Because a full-time worker with a bachelor’s degree earns $11 thousand more after taxes than a full-time worker with an associate’s degree, let

    x+11=the median yearly after-tax income of a full-time worker with a bachelors degree.

    Because a full-time worker with a master’s degree earns $21 thousand more after taxes than a full-time worker with an associate’s degree, let

    x+21=the median yearly after-tax income of a full-time worker with a master's degree.
  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. Combined, three full-time workers with each of these degrees earn $149 thousand.

    An equation with variable x to model the conditions.

  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    x+(x+11)+(x+21)=149This is the equation that models the problems conditions.3x+32=149Remove parentheses, regroup, and combine like terms.3x=117Subtract 32 from both sides.x=39Divide both sides by 3.

    Because we isolated the variable in the model and obtained x=39,

    median after-tax income with an associates degree=x=39median after-tax income with a bachelors degree=x+11=39+11=50median after-tax income with a masters degree=x+21=39+21=60.

    Full-time workers with associate’s degrees earn $39 thousand per year after taxes, full-time workers with bachelor’s degrees earn $50 thousand per year after taxes, and full-time workers with master’s degrees earn $60 thousand per year after taxes.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that combined, three full-time workers with each of these educational attainments earn $149 thousand after taxes. Using the incomes we determined in step 4, the sum is

    $39 thousand+$50 thousand+$60 thousand,or $149 thousand,

    which satisfies the problem’s conditions.

Check Point 1

  • The median yearly before-tax income of a full-time worker with a bachelor’s degree exceeds that of a full-time worker with an associate’s degree by $15 thousand. The median yearly before-tax income of a full-time worker with a master’s degree exceeds that of a full-time worker with an associate’s degree by $30 thousand. Combined, three full-time workers with each of these educational attainments earn $195 thousand before taxes. Find the median yearly before-tax income of full-time workers with each of these levels of education. (These incomes are illustrated by the bar graph on the previous page.)

Example 2 Modeling Attitudes of College Freshmen

Your author teaching math in 1969

Researchers have surveyed college freshmen every year since 1969. Figure P.15 shows that attitudes about some life goals have changed dramatically over the years. In particular, the freshman class of 2018 was more interested in making money than the freshmen of 1969 had been. In 1969, 42% of first-year college students considered “being well-off financially” essential or very important. For the period from 1969 through 2018, this percentage increased by approximately 0.8 each year. If this trend continues, by which year will all college freshmen consider “being well-off financially” essential or very important?

Figure P.15

A bar chart shows the life objectives of college freshmen, for the period from 19 69 to 20 18.

Source: Higher Education Research Institute

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We are interested in the year when all college freshmen, or 100% of the freshmen, will consider this life objective essential or very important. Let

    x = the number of years after 1969 when all freshmen will considerbeing    well-off financially essential or very important.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. There are no other unknown quantities to find, so we can skip this step.

  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS.

    The image shows a mathematical expression 42 + 0.8 x = 100
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    42+0.8x=100This is the equation that models the problems conditions.4242+0.8x=10042Subtract 42 from both sides.0.8x=58Simplify.0.8x0.8=580.8Divide both sides by 0.8.x=72.573Simplify and round to the nearestwhole number.

    Using current trends, by approximately 73 years after 1969, or in 2042, all freshmen will consider “being well-off financially” essential or very important.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that all freshmen (100%, represented by 100 using the model) will consider the objective essential or very important. Does this approximately occur if we increase the 1969 percentage, 42, by 0.8 each year for 73 years, our proposed solution?

    42+0.8(73)=42+58.4=100.4100

    This verifies that using trends shown in Figure P.15, all first-year college students will consider the objective of being well-off financially essential or very important approximately 73 years after 1969, or in 2042.

Check Point 2

  • Figure P.15 on the previous page shows that the freshman class of 2018 was less interested in developing a philosophy of life than the freshmen of 1969 had been. In 1969, 85% of the freshmen considered this objective essential or very important. Since then, this percentage has decreased by approximately 0.8 each year. If this trend continues, by which year will only 25% of college freshmen consider “developing a meaningful philosophy of life” essential or very important?

Example 3 Modeling Options for a Toll

It’s time to decide how you’re going to pay the bridge toll so that you can get to the beach. The first option requires purchasing a transponder for $20; with the transponder, you pay a reduced toll of $3.25 each time you cross the bridge. Your second option is toll-by-plate; with this option, you pay the full toll of $4.25 each time you cross the bridge plus a $3 administrative fee for each crossing. Find the number of times you would need to cross the bridge for the costs of the two options to be the same.

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. Let

    x=the number of times you cross the bridge.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. There are no other unknown quantities, so we can skip this step.

  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. The cost with the transponder is the cost of the transponder, $20, plus the toll, $3.25, times the number of crossings, x. For toll-by-plate, you must pay both the toll, $4.25, and the administrative fee, $3, each time you cross the bridge. The cost for toll-by-plate is the sum of the toll and the administrative fee, $4.25+$3 or $7.25, times the number of crossings, x. We want to know when these two costs are equal.

    The image shows a mathematical expression 7.25x = 20 + 3.25 x
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    7.25x=20+3.25xThis is the equation that models theproblems conditions.4x=20Subtract 3.25x from both sides.4x=5Divide both sides by 4.

    Because x represents the number of times you cross the bridge, the total cost of toll-by-plate is the same as the total cost with the transponder for 5 crossings.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that the cost of toll-by-plate should be the same as the cost with the transponder. Let’s see if they are the same with 5 bridge crossings per month.

    Expression to check the proposed solution.

    If you cross the bridge five times, both options cost $36.25. Thus the proposed solution, 5 crossings, satisfies the problem’s conditions.

Check Point 3

  • You drive up to a toll plaza and find booths with attendants where you can pay the toll by cash or credit card. With this option, the toll is $5 each time you cross the bridge. The attendant gives you the option of buying a bar-coded decal for $25; with the decal, you get 25% off the normal toll of $5 for each crossing. Find the number of times you would need to cross the bridge for the costs of the two options to be the same.

Example 4 A Price Reduction on Wireless Headphones

Your favorite electronics retailer is having a terrific sale on noise cancelling wireless headphones. (You’ll finally be able to listen to music in peace!) After a 40% price reduction, you purchase headphones for $192. What was the price of the headphones before the reduction?

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the price of the headphones prior to the reduction.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. There are no other unknown quantities to find, so we can skip this step.

  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. The headphones’ original price minus the 40% reduction is the reduced price, $192.

    The image shows a mathematical expression x minus 0.4 x = 192.
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    The solution of an equation.

    The headphones’ price before the reduction was $320.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The price before the reduction, $320, minus the 40% reduction should equal the reduced price given in the original wording, $192:

    32040% of 320=3200.4(320)=320128=192.

    This verifies that the headphones’ price before the reduction was $320.

Check Point 4

  • After a 30% price reduction, you purchase a new laptop for $840. What was the laptop’s price before the reduction?

Our next example is about simple interest. Simple interest involves interest calculated only on the amount of money that we invest, called the principal. The formula I=Pr is used to find the simple interest, I, earned for one year when the principal, P, is invested at an annual interest rate, r. Dual investment problems involve different amounts of money in two or more investments, each paying a different rate.

Example 5 Solving a Dual Investment Problem

Your grandmother needs your help. She has $150,000 to invest. Part of this money is to be invested in a money market account paying 1.5% annual interest. The rest of this money is to be invested in a certificate of deposit paying 1.3% annual interest. She told you that she requires $2000 per year in extra income from the combination of these investments. How much money should be placed in each investment?

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the amount invested in the money market account at 1.5%.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. The other quantity that we seek is the amount invested at 1.3% in the certificate of deposit. Because the total amount Grandma has to invest is $150,000 and we already used up x,

    150,000x=the amount invested in the certificate of deposit at 1.3%.
  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. Because Grandma requires $2000 in total interest, the interest for the two investments combined must be $2000. Interest is Pr or rP for each investment.

    The image shows a mathematical expression 0.015 x + 0.013 left parenthesis 150,000 minus x right parenthesis = 20 00
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    A mathematical expression steps to solve an equation to find the value of x.

    Thus,

    the amount invested at1.5%=x=25,000the amount invested at 1.3%=150,00025,000=125,000

    Grandma should invest $25,000 at 1.5% and $125,000 at 1.3%.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that the total interest from the dual investments should be $2000. Can Grandma count on $2000 interest? The interest earned on $25,000 at 1.5% is ($25,000) (0.015), or $375. The interest earned on $125,000 at 1.3% is ($125,000)(0.013), or $1625. The total interest is $375+$1625, or $2000, exactly as it should be. You’ve made your grandmother happy. (Now if you would just visit her more often …)

Check Point 5

  • You inherited $50,000 with the stipulation that for the first year the money had to be invested in two accounts paying 0.9% and 1.1% annual interest. How much did you invest at each rate if the total interest earned for the year was $515?

Solving geometry problems usually requires a knowledge of basic geometric ideas and formulas. Formulas for area, perimeter, and volume are given in Table P.6.

Table P.6 Common Formulas for Area, Perimeter, and Volume

An illustration shows common formulas for area, perimeter, and volume of different shapes.
Table P.6 Full Alternative Text

We will be using the formula for the perimeter of a rectangle, P=2l+2w, in our next example. The formula states that a rectangle’s perimeter is the sum of twice its length and twice its width.

Example 6 Finding the Dimensions of an American Football Field

The length of an American football field is 200 feet more than the width. If the perimeter of the field is 1040 feet, what are its dimensions?

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We know something about the length; the length is 200 feet more than the width. We will let

    x=the width.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. Because the length is 200 feet more than the width, we add 200 to the width to represent the length. Thus,

    x+200=the length.

    Figure P.16 illustrates an American football field and its dimensions.

    Figure P.16 An American football field

    An illustration shows an American football field with its width as x and length as x + 20 0.
  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. Because the perimeter of the field is 1040 feet,

    The image shows the mathematical expression 2 left parenthesis x + 20 0 right parenthesis + 2 x = 1040.

  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    Steps to solve an equation.

    Thus,

    width=x=160.length=x+200=160+200=360.

    The dimensions of an American football field are 160 feet by 360 feet. (The 360-foot length is usually described as 120 yards.)

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The perimeter of the football field using the dimensions that we found is

    2(160 feet)+2(360 feet)=320 feet+720 feet=1040 feet.

    Because the problem’s wording tells us that the perimeter is 1040 feet, our dimensions are correct.

Check Point 6

  • The length of a rectangular basketball court is 44 feet more than the width. If the perimeter of the basketball court is 288 feet, what are its dimensions?

We will use the formula for the area of a rectangle, A=lw, in our next example. The formula states that a rectangle’s area is the product of its length and its width.

Example 7 Solving a Problem Involving Landscape Design

A rectangular garden measures 80 feet by 60 feet. A large path of uniform width is to be added along both shorter sides and one longer side of the garden. The landscape designer doing the work wants to double the garden’s area with the addition of this path. How wide should the path be?

Solution

  1. Step 1 LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the width of the path.

    The situation is illustrated in Figure P.17. The figure shows the original 80-by-60 foot rectangular garden and the path of width x added along both shorter sides and one longer side.

    Figure P.17 The garden’s area is to be doubled by adding the path.

    An image shows a rectangular garden is surrounded by a path that has a length of 80 + 2 x and a width of 60 + x.
  2. Step 2 REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. Because the path is added along both shorter sides and one longer side, Figure P.17 shows that

     80+2x=the length of the new, expanded rectangle 60+x=the width of the new, expanded rectangle.

  3. Step 3 WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. The area of the rectangle must be doubled by the addition of the path.

    Left parenthesis 80 + 2 x right parenthesis, left parenthesis 60 + x right parenthesis = 2 times 80 times 60.
  4. Step 4 SOLVE THE EQUATION AND ANSWER THE QUESTION.

    (80+2x)(60+x)=28060This is the equation that modelsthe problems conditions.4800+200x+2x2=9600Multiply. Use FOIL on the left side.2x2+200x4800=0Subtract 9600 from both sidesand write the quadratic equationin standard form.2(x2+100x2400)=0Factor out 2, the GCF.2(x20)(x+120)=0Factor the trinomial.x20=0or x+120=0Set each variable factor equal to 0.x=20x=120Solve for x.

    The path cannot have a negative width. Because 120 is geometrically impossible, we use x=20. The width of the path should be 20 feet.

  5. Step 5 CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. Has the landscape architect doubled the garden’s area with the 20-foot-wide path? The area of the garden is 80 feet times 60 feet, or 4800 square feet. Because 80+2x and 60+x represent the length and width of the expanded rectangle,

    80+2x=80+220 =120 feet is the expanded rectangle's length.60+2x=60+20=80 feet is the expanded rectangle's width.

    The area of the expanded rectangle is 120 feet times 80 feet, or 9600 square feet. This is double the area of the garden, 4800 square feet, as specified by the problem’s conditions.

Check Point 7

  • A rectangular garden measures 16 feet by 12 feet. A path of uniform width is to be added so as to surround the entire garden. The landscape artist doing the work wants the garden and path to cover an area of 320 square feet. How wide should the path be?

In our next example, we will be using the Pythagorean Theorem to obtain a mathematical model. The ancient Greek philosopher and mathematician Pythagoras (approximately 582–500 b.c.) founded a school whose motto was “All is number.” Pythagoras is best remembered for his work with the right triangle, a triangle with one angle measuring 90°. The side opposite the 90° angle is called the hypotenuse. The other sides are called legs. Pythagoras found that if he constructed squares on each of the legs, as well as a larger square on the hypotenuse, the sum of the areas of the smaller squares is equal to the area of the larger square. This is illustrated in Figure P.18.

Figure P.18The area of the large square equals the sum of the areas of the smaller squares.

An illustration shows a right triangle of base 3, height 4, and hypotenuse 5.
Figure P. 18 Full Alternative Text

This relationship is usually stated in terms of the lengths of the three sides of a right triangle and is called the Pythagorean Theorem.

The Pythagorean Theorem

The sum of the squares of the lengths of the legs of a right triangle equals the square of the length of the hypotenuse.

If the legs have lengths a and b, and the hypotenuse has length c, then

a2+b2=c2.
An image shows a right angle triangle ABC, which represents angle C is a 90 degree angle. AC represents as a base labeled, b, Leg, BC denotes as a height labeled, a Leg, and AB is labeled, c, hypotenuse.

Example 8 Using the Pythagorean Theorem: Screen Math

A bar graph shows the rising percentage of large screen TVs on store shelves.

Source: GAP Intelligence

Did you know that the size of a television screen refers to the length of its diagonal? If the length of the HDTV screen in Figure P.19 is 28 inches and its width is 15.7 inches, what is the size of the screen to the nearest inch?

Figure P.19

A photo shows a television with a screen size of length 28 inches and a width of 15.7 inches.

Solution

Figure P.19 shows that the length, width, and diagonal of the screen form a right triangle. The diagonal is the hypotenuse of the triangle. We use the Pythagorean Theorem with a=28, b=15.7, and solve for the screen size, c.

a2+b2=c2This is the Pythagorean Theorem.282+15.72=c2Let a=28 and b=15.7.784+246.49=c2282=2828=784 and15.72=15.715.7=246.49.c2=1030.49Add: 784+246.49=1030.49. We also reversedthe two sides.
c=1030.49 or  c=1030.49Apply the square root property.c32 or  c32Use a calculator and round to thenearest inch. 1030.49    = or   1030.49 ENTER

Because c represents the size of the screen, this dimension must be positive. We reject 32. Thus, the screen size of the HDTV is 32 inches.

Check Point 8

  • Figure P.20 shows the dimensions of an old TV screen. What is the size of the screen?

    Figure P.20

    A photo shows a television with a screen size of length 25.6 inches and a width of 19.2 inches.

Example 9 Dividing the Cost of a Yacht

A group of friends agrees to share the cost of a $50,000 yacht equally. Before the purchase is made, one more person joins the group. As a result, each person’s share is reduced by $2500. How many people were in the original group?

Solution

  1. Step 1 LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the number of people in the original group.
  2. Step 2 REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x.  Because one more person joined the original group, let

    x+1=the number of people in the final group.
  3. Step 3 WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. As a result of one more person joining the original group, each person’s share is reduced by $2500.

    The image shows a mathematical expression, start fraction 50,000 over x end fraction minus 2500 = start fraction 50,000 over x + 1 end fraction.
  4. Step 4 SOLVE THE EQUATION AND ANSWER THE QUESTION.

    50,000x2500=50,000x+1This is the equation that models theproblems conditions.x(x+1)(50,000x2500)=x(x+1)50,000x+1Multiply both sides by x(x+1),the LCD. x (x+1)50,000 x x(x+1)2500=x (x+1) 50,000 (x+1) Use the distributive property anddivide out common factors.50,000(x+1)2500x(x+1)=50,000xSimplify.50,000x+50,0002500x22500x=50,000xUse the distributive property.2500x2+47,500x+50,000=50,000xCombine like terms.2500x22500x+50,000=0Write the quadratic equation instandard form.2500(x2+x20)=0Factor out 2500.2500(x+5)(x4)=0Factor completely.x+5=0or x4=0Set each variable factor equal to zero.x=5x=4Solve the resulting equations.

    Because x represents the number of people in the original group, x cannot be negative. Thus, there were four people in the original group.

  5. Step 5 CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM.

     original cost per person=$50,0004=$12,500 final cost per person=$50,0005=$10,000

    We see that each person’s share is reduced by $12,500$10,000, or $2500, as specified by the problem’s conditions.

Check Point 9

  • A group of people share equally in a $5,000,000 lottery. Before the money is divided, three more winning ticket holders are declared. As a result, each person’s share is reduced by $375,000. How many people were in the original group of winners?

Objective 1: Use equations to solve problems

Problem Solving with Equations

  1. Objective 1Use equations to solve problems.

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We have seen that a model is a mathematical representation of a real-world situation. In this section, we will be solving problems that are presented in English. This means that we must obtain models by translating from the ordinary language of English into the language of algebraic equations. To translate, however, we must understand the English prose and be familiar with the forms of algebraic language. Following are some general steps we will use in solving word problems:

Strategy for Solving Word Problems

  1. Step 1 Read the problem carefully several times until you can state in your own words what is given and what the problem is looking for. Let x (or any variable) represent one of the unknown quantities in the problem.

  2. Step 2 If necessary, write expressions for any other unknown quantities in the problem in terms of x.

  3. Step 3 Write an equation in x that models the verbal conditions of the problem.

  4. Step 4 Solve the equation and answer the problem’s question.

  5. Step 5 Check the solution in the original wording of the problem, not in the equation obtained from the words.

Example 1 Education Pays Off

The graph in Figure P.14 shows median yearly earnings in the United States by highest educational attainment. It certainly looks like you can make more money with more education. However, workers who make more money pay a higher percentage of income in taxes, so it is wise to consider after-tax earnings when deciding if that higher degree will pay off.

Figure P.14

A bar chart shows median earnings of full-time workers in the United States, by highest educational attainment.

Source: Education Pays 2019

The median yearly after-tax income of a full-time worker with a bachelor’s degree exceeds that of a full-time worker with an associate’s degree by $11 thousand. The median yearly after-tax income of a full-time worker with a master’s degree exceeds that of a full-time worker with an associate’s degree by $21 thousand. Combined, three full-time workers with each of these degrees earn $149 thousand after taxes. Find the median yearly after-tax income of full-time workers with each of these levels of education.

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We know something about after-tax incomes of full-time workers with bachelor’s degrees and master’s degrees: They exceed the after-tax income of a full-time worker with an associate’s degree by $11 thousand and $21 thousand, respectively. We will let

    x=the median yearly after-tax income of a full-time worker with an associates degree.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. Because a full-time worker with a bachelor’s degree earns $11 thousand more after taxes than a full-time worker with an associate’s degree, let

    x+11=the median yearly after-tax income of a full-time worker with a bachelors degree.

    Because a full-time worker with a master’s degree earns $21 thousand more after taxes than a full-time worker with an associate’s degree, let

    x+21=the median yearly after-tax income of a full-time worker with a master's degree.
  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. Combined, three full-time workers with each of these degrees earn $149 thousand.

    An equation with variable x to model the conditions.

  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    x+(x+11)+(x+21)=149This is the equation that models the problems conditions.3x+32=149Remove parentheses, regroup, and combine like terms.3x=117Subtract 32 from both sides.x=39Divide both sides by 3.

    Because we isolated the variable in the model and obtained x=39,

    median after-tax income with an associates degree=x=39median after-tax income with a bachelors degree=x+11=39+11=50median after-tax income with a masters degree=x+21=39+21=60.

    Full-time workers with associate’s degrees earn $39 thousand per year after taxes, full-time workers with bachelor’s degrees earn $50 thousand per year after taxes, and full-time workers with master’s degrees earn $60 thousand per year after taxes.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that combined, three full-time workers with each of these educational attainments earn $149 thousand after taxes. Using the incomes we determined in step 4, the sum is

    $39 thousand+$50 thousand+$60 thousand,or $149 thousand,

    which satisfies the problem’s conditions.

Check Point 1

  • The median yearly before-tax income of a full-time worker with a bachelor’s degree exceeds that of a full-time worker with an associate’s degree by $15 thousand. The median yearly before-tax income of a full-time worker with a master’s degree exceeds that of a full-time worker with an associate’s degree by $30 thousand. Combined, three full-time workers with each of these educational attainments earn $195 thousand before taxes. Find the median yearly before-tax income of full-time workers with each of these levels of education. (These incomes are illustrated by the bar graph on the previous page.)

Example 2 Modeling Attitudes of College Freshmen

Your author teaching math in 1969

Researchers have surveyed college freshmen every year since 1969. Figure P.15 shows that attitudes about some life goals have changed dramatically over the years. In particular, the freshman class of 2018 was more interested in making money than the freshmen of 1969 had been. In 1969, 42% of first-year college students considered “being well-off financially” essential or very important. For the period from 1969 through 2018, this percentage increased by approximately 0.8 each year. If this trend continues, by which year will all college freshmen consider “being well-off financially” essential or very important?

Figure P.15

A bar chart shows the life objectives of college freshmen, for the period from 19 69 to 20 18.

Source: Higher Education Research Institute

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We are interested in the year when all college freshmen, or 100% of the freshmen, will consider this life objective essential or very important. Let

    x = the number of years after 1969 when all freshmen will considerbeing    well-off financially essential or very important.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. There are no other unknown quantities to find, so we can skip this step.

  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS.

    The image shows a mathematical expression 42 + 0.8 x = 100
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    42+0.8x=100This is the equation that models the problems conditions.4242+0.8x=10042Subtract 42 from both sides.0.8x=58Simplify.0.8x0.8=580.8Divide both sides by 0.8.x=72.573Simplify and round to the nearestwhole number.

    Using current trends, by approximately 73 years after 1969, or in 2042, all freshmen will consider “being well-off financially” essential or very important.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that all freshmen (100%, represented by 100 using the model) will consider the objective essential or very important. Does this approximately occur if we increase the 1969 percentage, 42, by 0.8 each year for 73 years, our proposed solution?

    42+0.8(73)=42+58.4=100.4100

    This verifies that using trends shown in Figure P.15, all first-year college students will consider the objective of being well-off financially essential or very important approximately 73 years after 1969, or in 2042.

Check Point 2

  • Figure P.15 on the previous page shows that the freshman class of 2018 was less interested in developing a philosophy of life than the freshmen of 1969 had been. In 1969, 85% of the freshmen considered this objective essential or very important. Since then, this percentage has decreased by approximately 0.8 each year. If this trend continues, by which year will only 25% of college freshmen consider “developing a meaningful philosophy of life” essential or very important?

Example 3 Modeling Options for a Toll

It’s time to decide how you’re going to pay the bridge toll so that you can get to the beach. The first option requires purchasing a transponder for $20; with the transponder, you pay a reduced toll of $3.25 each time you cross the bridge. Your second option is toll-by-plate; with this option, you pay the full toll of $4.25 each time you cross the bridge plus a $3 administrative fee for each crossing. Find the number of times you would need to cross the bridge for the costs of the two options to be the same.

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. Let

    x=the number of times you cross the bridge.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. There are no other unknown quantities, so we can skip this step.

  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. The cost with the transponder is the cost of the transponder, $20, plus the toll, $3.25, times the number of crossings, x. For toll-by-plate, you must pay both the toll, $4.25, and the administrative fee, $3, each time you cross the bridge. The cost for toll-by-plate is the sum of the toll and the administrative fee, $4.25+$3 or $7.25, times the number of crossings, x. We want to know when these two costs are equal.

    The image shows a mathematical expression 7.25x = 20 + 3.25 x
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    7.25x=20+3.25xThis is the equation that models theproblems conditions.4x=20Subtract 3.25x from both sides.4x=5Divide both sides by 4.

    Because x represents the number of times you cross the bridge, the total cost of toll-by-plate is the same as the total cost with the transponder for 5 crossings.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that the cost of toll-by-plate should be the same as the cost with the transponder. Let’s see if they are the same with 5 bridge crossings per month.

    Expression to check the proposed solution.

    If you cross the bridge five times, both options cost $36.25. Thus the proposed solution, 5 crossings, satisfies the problem’s conditions.

Check Point 3

  • You drive up to a toll plaza and find booths with attendants where you can pay the toll by cash or credit card. With this option, the toll is $5 each time you cross the bridge. The attendant gives you the option of buying a bar-coded decal for $25; with the decal, you get 25% off the normal toll of $5 for each crossing. Find the number of times you would need to cross the bridge for the costs of the two options to be the same.

Example 4 A Price Reduction on Wireless Headphones

Your favorite electronics retailer is having a terrific sale on noise cancelling wireless headphones. (You’ll finally be able to listen to music in peace!) After a 40% price reduction, you purchase headphones for $192. What was the price of the headphones before the reduction?

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the price of the headphones prior to the reduction.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. There are no other unknown quantities to find, so we can skip this step.

  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. The headphones’ original price minus the 40% reduction is the reduced price, $192.

    The image shows a mathematical expression x minus 0.4 x = 192.
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    The solution of an equation.

    The headphones’ price before the reduction was $320.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The price before the reduction, $320, minus the 40% reduction should equal the reduced price given in the original wording, $192:

    32040% of 320=3200.4(320)=320128=192.

    This verifies that the headphones’ price before the reduction was $320.

Check Point 4

  • After a 30% price reduction, you purchase a new laptop for $840. What was the laptop’s price before the reduction?

Our next example is about simple interest. Simple interest involves interest calculated only on the amount of money that we invest, called the principal. The formula I=Pr is used to find the simple interest, I, earned for one year when the principal, P, is invested at an annual interest rate, r. Dual investment problems involve different amounts of money in two or more investments, each paying a different rate.

Example 5 Solving a Dual Investment Problem

Your grandmother needs your help. She has $150,000 to invest. Part of this money is to be invested in a money market account paying 1.5% annual interest. The rest of this money is to be invested in a certificate of deposit paying 1.3% annual interest. She told you that she requires $2000 per year in extra income from the combination of these investments. How much money should be placed in each investment?

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the amount invested in the money market account at 1.5%.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. The other quantity that we seek is the amount invested at 1.3% in the certificate of deposit. Because the total amount Grandma has to invest is $150,000 and we already used up x,

    150,000x=the amount invested in the certificate of deposit at 1.3%.
  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. Because Grandma requires $2000 in total interest, the interest for the two investments combined must be $2000. Interest is Pr or rP for each investment.

    The image shows a mathematical expression 0.015 x + 0.013 left parenthesis 150,000 minus x right parenthesis = 20 00
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    A mathematical expression steps to solve an equation to find the value of x.

    Thus,

    the amount invested at1.5%=x=25,000the amount invested at 1.3%=150,00025,000=125,000

    Grandma should invest $25,000 at 1.5% and $125,000 at 1.3%.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that the total interest from the dual investments should be $2000. Can Grandma count on $2000 interest? The interest earned on $25,000 at 1.5% is ($25,000) (0.015), or $375. The interest earned on $125,000 at 1.3% is ($125,000)(0.013), or $1625. The total interest is $375+$1625, or $2000, exactly as it should be. You’ve made your grandmother happy. (Now if you would just visit her more often …)

Check Point 5

  • You inherited $50,000 with the stipulation that for the first year the money had to be invested in two accounts paying 0.9% and 1.1% annual interest. How much did you invest at each rate if the total interest earned for the year was $515?

Solving geometry problems usually requires a knowledge of basic geometric ideas and formulas. Formulas for area, perimeter, and volume are given in Table P.6.

Table P.6 Common Formulas for Area, Perimeter, and Volume

An illustration shows common formulas for area, perimeter, and volume of different shapes.
Table P.6 Full Alternative Text

We will be using the formula for the perimeter of a rectangle, P=2l+2w, in our next example. The formula states that a rectangle’s perimeter is the sum of twice its length and twice its width.

Example 6 Finding the Dimensions of an American Football Field

The length of an American football field is 200 feet more than the width. If the perimeter of the field is 1040 feet, what are its dimensions?

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We know something about the length; the length is 200 feet more than the width. We will let

    x=the width.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. Because the length is 200 feet more than the width, we add 200 to the width to represent the length. Thus,

    x+200=the length.

    Figure P.16 illustrates an American football field and its dimensions.

    Figure P.16 An American football field

    An illustration shows an American football field with its width as x and length as x + 20 0.
  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. Because the perimeter of the field is 1040 feet,

    The image shows the mathematical expression 2 left parenthesis x + 20 0 right parenthesis + 2 x = 1040.

  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    Steps to solve an equation.

    Thus,

    width=x=160.length=x+200=160+200=360.

    The dimensions of an American football field are 160 feet by 360 feet. (The 360-foot length is usually described as 120 yards.)

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The perimeter of the football field using the dimensions that we found is

    2(160 feet)+2(360 feet)=320 feet+720 feet=1040 feet.

    Because the problem’s wording tells us that the perimeter is 1040 feet, our dimensions are correct.

Check Point 6

  • The length of a rectangular basketball court is 44 feet more than the width. If the perimeter of the basketball court is 288 feet, what are its dimensions?

We will use the formula for the area of a rectangle, A=lw, in our next example. The formula states that a rectangle’s area is the product of its length and its width.

Example 7 Solving a Problem Involving Landscape Design

A rectangular garden measures 80 feet by 60 feet. A large path of uniform width is to be added along both shorter sides and one longer side of the garden. The landscape designer doing the work wants to double the garden’s area with the addition of this path. How wide should the path be?

Solution

  1. Step 1 LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the width of the path.

    The situation is illustrated in Figure P.17. The figure shows the original 80-by-60 foot rectangular garden and the path of width x added along both shorter sides and one longer side.

    Figure P.17 The garden’s area is to be doubled by adding the path.

    An image shows a rectangular garden is surrounded by a path that has a length of 80 + 2 x and a width of 60 + x.
  2. Step 2 REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. Because the path is added along both shorter sides and one longer side, Figure P.17 shows that

     80+2x=the length of the new, expanded rectangle 60+x=the width of the new, expanded rectangle.

  3. Step 3 WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. The area of the rectangle must be doubled by the addition of the path.

    Left parenthesis 80 + 2 x right parenthesis, left parenthesis 60 + x right parenthesis = 2 times 80 times 60.
  4. Step 4 SOLVE THE EQUATION AND ANSWER THE QUESTION.

    (80+2x)(60+x)=28060This is the equation that modelsthe problems conditions.4800+200x+2x2=9600Multiply. Use FOIL on the left side.2x2+200x4800=0Subtract 9600 from both sidesand write the quadratic equationin standard form.2(x2+100x2400)=0Factor out 2, the GCF.2(x20)(x+120)=0Factor the trinomial.x20=0or x+120=0Set each variable factor equal to 0.x=20x=120Solve for x.

    The path cannot have a negative width. Because 120 is geometrically impossible, we use x=20. The width of the path should be 20 feet.

  5. Step 5 CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. Has the landscape architect doubled the garden’s area with the 20-foot-wide path? The area of the garden is 80 feet times 60 feet, or 4800 square feet. Because 80+2x and 60+x represent the length and width of the expanded rectangle,

    80+2x=80+220 =120 feet is the expanded rectangle's length.60+2x=60+20=80 feet is the expanded rectangle's width.

    The area of the expanded rectangle is 120 feet times 80 feet, or 9600 square feet. This is double the area of the garden, 4800 square feet, as specified by the problem’s conditions.

Check Point 7

  • A rectangular garden measures 16 feet by 12 feet. A path of uniform width is to be added so as to surround the entire garden. The landscape artist doing the work wants the garden and path to cover an area of 320 square feet. How wide should the path be?

In our next example, we will be using the Pythagorean Theorem to obtain a mathematical model. The ancient Greek philosopher and mathematician Pythagoras (approximately 582–500 b.c.) founded a school whose motto was “All is number.” Pythagoras is best remembered for his work with the right triangle, a triangle with one angle measuring 90°. The side opposite the 90° angle is called the hypotenuse. The other sides are called legs. Pythagoras found that if he constructed squares on each of the legs, as well as a larger square on the hypotenuse, the sum of the areas of the smaller squares is equal to the area of the larger square. This is illustrated in Figure P.18.

Figure P.18The area of the large square equals the sum of the areas of the smaller squares.

An illustration shows a right triangle of base 3, height 4, and hypotenuse 5.
Figure P. 18 Full Alternative Text

This relationship is usually stated in terms of the lengths of the three sides of a right triangle and is called the Pythagorean Theorem.

The Pythagorean Theorem

The sum of the squares of the lengths of the legs of a right triangle equals the square of the length of the hypotenuse.

If the legs have lengths a and b, and the hypotenuse has length c, then

a2+b2=c2.
An image shows a right angle triangle ABC, which represents angle C is a 90 degree angle. AC represents as a base labeled, b, Leg, BC denotes as a height labeled, a Leg, and AB is labeled, c, hypotenuse.

Example 8 Using the Pythagorean Theorem: Screen Math

A bar graph shows the rising percentage of large screen TVs on store shelves.

Source: GAP Intelligence

Did you know that the size of a television screen refers to the length of its diagonal? If the length of the HDTV screen in Figure P.19 is 28 inches and its width is 15.7 inches, what is the size of the screen to the nearest inch?

Figure P.19

A photo shows a television with a screen size of length 28 inches and a width of 15.7 inches.

Solution

Figure P.19 shows that the length, width, and diagonal of the screen form a right triangle. The diagonal is the hypotenuse of the triangle. We use the Pythagorean Theorem with a=28, b=15.7, and solve for the screen size, c.

a2+b2=c2This is the Pythagorean Theorem.282+15.72=c2Let a=28 and b=15.7.784+246.49=c2282=2828=784 and15.72=15.715.7=246.49.c2=1030.49Add: 784+246.49=1030.49. We also reversedthe two sides.
c=1030.49 or  c=1030.49Apply the square root property.c32 or  c32Use a calculator and round to thenearest inch. 1030.49    = or   1030.49 ENTER

Because c represents the size of the screen, this dimension must be positive. We reject 32. Thus, the screen size of the HDTV is 32 inches.

Check Point 8

  • Figure P.20 shows the dimensions of an old TV screen. What is the size of the screen?

    Figure P.20

    A photo shows a television with a screen size of length 25.6 inches and a width of 19.2 inches.

Example 9 Dividing the Cost of a Yacht

A group of friends agrees to share the cost of a $50,000 yacht equally. Before the purchase is made, one more person joins the group. As a result, each person’s share is reduced by $2500. How many people were in the original group?

Solution

  1. Step 1 LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the number of people in the original group.
  2. Step 2 REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x.  Because one more person joined the original group, let

    x+1=the number of people in the final group.
  3. Step 3 WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. As a result of one more person joining the original group, each person’s share is reduced by $2500.

    The image shows a mathematical expression, start fraction 50,000 over x end fraction minus 2500 = start fraction 50,000 over x + 1 end fraction.
  4. Step 4 SOLVE THE EQUATION AND ANSWER THE QUESTION.

    50,000x2500=50,000x+1This is the equation that models theproblems conditions.x(x+1)(50,000x2500)=x(x+1)50,000x+1Multiply both sides by x(x+1),the LCD. x (x+1)50,000 x x(x+1)2500=x (x+1) 50,000 (x+1) Use the distributive property anddivide out common factors.50,000(x+1)2500x(x+1)=50,000xSimplify.50,000x+50,0002500x22500x=50,000xUse the distributive property.2500x2+47,500x+50,000=50,000xCombine like terms.2500x22500x+50,000=0Write the quadratic equation instandard form.2500(x2+x20)=0Factor out 2500.2500(x+5)(x4)=0Factor completely.x+5=0or x4=0Set each variable factor equal to zero.x=5x=4Solve the resulting equations.

    Because x represents the number of people in the original group, x cannot be negative. Thus, there were four people in the original group.

  5. Step 5 CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM.

     original cost per person=$50,0004=$12,500 final cost per person=$50,0005=$10,000

    We see that each person’s share is reduced by $12,500$10,000, or $2500, as specified by the problem’s conditions.

Check Point 9

  • A group of people share equally in a $5,000,000 lottery. Before the money is divided, three more winning ticket holders are declared. As a result, each person’s share is reduced by $375,000. How many people were in the original group of winners?

Objective 1: Use equations to solve problems

Problem Solving with Equations

  1. Objective 1Use equations to solve problems.

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We have seen that a model is a mathematical representation of a real-world situation. In this section, we will be solving problems that are presented in English. This means that we must obtain models by translating from the ordinary language of English into the language of algebraic equations. To translate, however, we must understand the English prose and be familiar with the forms of algebraic language. Following are some general steps we will use in solving word problems:

Strategy for Solving Word Problems

  1. Step 1 Read the problem carefully several times until you can state in your own words what is given and what the problem is looking for. Let x (or any variable) represent one of the unknown quantities in the problem.

  2. Step 2 If necessary, write expressions for any other unknown quantities in the problem in terms of x.

  3. Step 3 Write an equation in x that models the verbal conditions of the problem.

  4. Step 4 Solve the equation and answer the problem’s question.

  5. Step 5 Check the solution in the original wording of the problem, not in the equation obtained from the words.

Example 1 Education Pays Off

The graph in Figure P.14 shows median yearly earnings in the United States by highest educational attainment. It certainly looks like you can make more money with more education. However, workers who make more money pay a higher percentage of income in taxes, so it is wise to consider after-tax earnings when deciding if that higher degree will pay off.

Figure P.14

A bar chart shows median earnings of full-time workers in the United States, by highest educational attainment.

Source: Education Pays 2019

The median yearly after-tax income of a full-time worker with a bachelor’s degree exceeds that of a full-time worker with an associate’s degree by $11 thousand. The median yearly after-tax income of a full-time worker with a master’s degree exceeds that of a full-time worker with an associate’s degree by $21 thousand. Combined, three full-time workers with each of these degrees earn $149 thousand after taxes. Find the median yearly after-tax income of full-time workers with each of these levels of education.

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We know something about after-tax incomes of full-time workers with bachelor’s degrees and master’s degrees: They exceed the after-tax income of a full-time worker with an associate’s degree by $11 thousand and $21 thousand, respectively. We will let

    x=the median yearly after-tax income of a full-time worker with an associates degree.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. Because a full-time worker with a bachelor’s degree earns $11 thousand more after taxes than a full-time worker with an associate’s degree, let

    x+11=the median yearly after-tax income of a full-time worker with a bachelors degree.

    Because a full-time worker with a master’s degree earns $21 thousand more after taxes than a full-time worker with an associate’s degree, let

    x+21=the median yearly after-tax income of a full-time worker with a master's degree.
  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. Combined, three full-time workers with each of these degrees earn $149 thousand.

    An equation with variable x to model the conditions.

  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    x+(x+11)+(x+21)=149This is the equation that models the problems conditions.3x+32=149Remove parentheses, regroup, and combine like terms.3x=117Subtract 32 from both sides.x=39Divide both sides by 3.

    Because we isolated the variable in the model and obtained x=39,

    median after-tax income with an associates degree=x=39median after-tax income with a bachelors degree=x+11=39+11=50median after-tax income with a masters degree=x+21=39+21=60.

    Full-time workers with associate’s degrees earn $39 thousand per year after taxes, full-time workers with bachelor’s degrees earn $50 thousand per year after taxes, and full-time workers with master’s degrees earn $60 thousand per year after taxes.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that combined, three full-time workers with each of these educational attainments earn $149 thousand after taxes. Using the incomes we determined in step 4, the sum is

    $39 thousand+$50 thousand+$60 thousand,or $149 thousand,

    which satisfies the problem’s conditions.

Check Point 1

  • The median yearly before-tax income of a full-time worker with a bachelor’s degree exceeds that of a full-time worker with an associate’s degree by $15 thousand. The median yearly before-tax income of a full-time worker with a master’s degree exceeds that of a full-time worker with an associate’s degree by $30 thousand. Combined, three full-time workers with each of these educational attainments earn $195 thousand before taxes. Find the median yearly before-tax income of full-time workers with each of these levels of education. (These incomes are illustrated by the bar graph on the previous page.)

Example 2 Modeling Attitudes of College Freshmen

Your author teaching math in 1969

Researchers have surveyed college freshmen every year since 1969. Figure P.15 shows that attitudes about some life goals have changed dramatically over the years. In particular, the freshman class of 2018 was more interested in making money than the freshmen of 1969 had been. In 1969, 42% of first-year college students considered “being well-off financially” essential or very important. For the period from 1969 through 2018, this percentage increased by approximately 0.8 each year. If this trend continues, by which year will all college freshmen consider “being well-off financially” essential or very important?

Figure P.15

A bar chart shows the life objectives of college freshmen, for the period from 19 69 to 20 18.

Source: Higher Education Research Institute

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We are interested in the year when all college freshmen, or 100% of the freshmen, will consider this life objective essential or very important. Let

    x = the number of years after 1969 when all freshmen will considerbeing    well-off financially essential or very important.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. There are no other unknown quantities to find, so we can skip this step.

  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS.

    The image shows a mathematical expression 42 + 0.8 x = 100
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    42+0.8x=100This is the equation that models the problems conditions.4242+0.8x=10042Subtract 42 from both sides.0.8x=58Simplify.0.8x0.8=580.8Divide both sides by 0.8.x=72.573Simplify and round to the nearestwhole number.

    Using current trends, by approximately 73 years after 1969, or in 2042, all freshmen will consider “being well-off financially” essential or very important.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that all freshmen (100%, represented by 100 using the model) will consider the objective essential or very important. Does this approximately occur if we increase the 1969 percentage, 42, by 0.8 each year for 73 years, our proposed solution?

    42+0.8(73)=42+58.4=100.4100

    This verifies that using trends shown in Figure P.15, all first-year college students will consider the objective of being well-off financially essential or very important approximately 73 years after 1969, or in 2042.

Check Point 2

  • Figure P.15 on the previous page shows that the freshman class of 2018 was less interested in developing a philosophy of life than the freshmen of 1969 had been. In 1969, 85% of the freshmen considered this objective essential or very important. Since then, this percentage has decreased by approximately 0.8 each year. If this trend continues, by which year will only 25% of college freshmen consider “developing a meaningful philosophy of life” essential or very important?

Example 3 Modeling Options for a Toll

It’s time to decide how you’re going to pay the bridge toll so that you can get to the beach. The first option requires purchasing a transponder for $20; with the transponder, you pay a reduced toll of $3.25 each time you cross the bridge. Your second option is toll-by-plate; with this option, you pay the full toll of $4.25 each time you cross the bridge plus a $3 administrative fee for each crossing. Find the number of times you would need to cross the bridge for the costs of the two options to be the same.

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. Let

    x=the number of times you cross the bridge.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. There are no other unknown quantities, so we can skip this step.

  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. The cost with the transponder is the cost of the transponder, $20, plus the toll, $3.25, times the number of crossings, x. For toll-by-plate, you must pay both the toll, $4.25, and the administrative fee, $3, each time you cross the bridge. The cost for toll-by-plate is the sum of the toll and the administrative fee, $4.25+$3 or $7.25, times the number of crossings, x. We want to know when these two costs are equal.

    The image shows a mathematical expression 7.25x = 20 + 3.25 x
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    7.25x=20+3.25xThis is the equation that models theproblems conditions.4x=20Subtract 3.25x from both sides.4x=5Divide both sides by 4.

    Because x represents the number of times you cross the bridge, the total cost of toll-by-plate is the same as the total cost with the transponder for 5 crossings.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that the cost of toll-by-plate should be the same as the cost with the transponder. Let’s see if they are the same with 5 bridge crossings per month.

    Expression to check the proposed solution.

    If you cross the bridge five times, both options cost $36.25. Thus the proposed solution, 5 crossings, satisfies the problem’s conditions.

Check Point 3

  • You drive up to a toll plaza and find booths with attendants where you can pay the toll by cash or credit card. With this option, the toll is $5 each time you cross the bridge. The attendant gives you the option of buying a bar-coded decal for $25; with the decal, you get 25% off the normal toll of $5 for each crossing. Find the number of times you would need to cross the bridge for the costs of the two options to be the same.

Example 4 A Price Reduction on Wireless Headphones

Your favorite electronics retailer is having a terrific sale on noise cancelling wireless headphones. (You’ll finally be able to listen to music in peace!) After a 40% price reduction, you purchase headphones for $192. What was the price of the headphones before the reduction?

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the price of the headphones prior to the reduction.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. There are no other unknown quantities to find, so we can skip this step.

  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. The headphones’ original price minus the 40% reduction is the reduced price, $192.

    The image shows a mathematical expression x minus 0.4 x = 192.
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    The solution of an equation.

    The headphones’ price before the reduction was $320.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The price before the reduction, $320, minus the 40% reduction should equal the reduced price given in the original wording, $192:

    32040% of 320=3200.4(320)=320128=192.

    This verifies that the headphones’ price before the reduction was $320.

Check Point 4

  • After a 30% price reduction, you purchase a new laptop for $840. What was the laptop’s price before the reduction?

Our next example is about simple interest. Simple interest involves interest calculated only on the amount of money that we invest, called the principal. The formula I=Pr is used to find the simple interest, I, earned for one year when the principal, P, is invested at an annual interest rate, r. Dual investment problems involve different amounts of money in two or more investments, each paying a different rate.

Example 5 Solving a Dual Investment Problem

Your grandmother needs your help. She has $150,000 to invest. Part of this money is to be invested in a money market account paying 1.5% annual interest. The rest of this money is to be invested in a certificate of deposit paying 1.3% annual interest. She told you that she requires $2000 per year in extra income from the combination of these investments. How much money should be placed in each investment?

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the amount invested in the money market account at 1.5%.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. The other quantity that we seek is the amount invested at 1.3% in the certificate of deposit. Because the total amount Grandma has to invest is $150,000 and we already used up x,

    150,000x=the amount invested in the certificate of deposit at 1.3%.
  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. Because Grandma requires $2000 in total interest, the interest for the two investments combined must be $2000. Interest is Pr or rP for each investment.

    The image shows a mathematical expression 0.015 x + 0.013 left parenthesis 150,000 minus x right parenthesis = 20 00
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    A mathematical expression steps to solve an equation to find the value of x.

    Thus,

    the amount invested at1.5%=x=25,000the amount invested at 1.3%=150,00025,000=125,000

    Grandma should invest $25,000 at 1.5% and $125,000 at 1.3%.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that the total interest from the dual investments should be $2000. Can Grandma count on $2000 interest? The interest earned on $25,000 at 1.5% is ($25,000) (0.015), or $375. The interest earned on $125,000 at 1.3% is ($125,000)(0.013), or $1625. The total interest is $375+$1625, or $2000, exactly as it should be. You’ve made your grandmother happy. (Now if you would just visit her more often …)

Check Point 5

  • You inherited $50,000 with the stipulation that for the first year the money had to be invested in two accounts paying 0.9% and 1.1% annual interest. How much did you invest at each rate if the total interest earned for the year was $515?

Solving geometry problems usually requires a knowledge of basic geometric ideas and formulas. Formulas for area, perimeter, and volume are given in Table P.6.

Table P.6 Common Formulas for Area, Perimeter, and Volume

An illustration shows common formulas for area, perimeter, and volume of different shapes.
Table P.6 Full Alternative Text

We will be using the formula for the perimeter of a rectangle, P=2l+2w, in our next example. The formula states that a rectangle’s perimeter is the sum of twice its length and twice its width.

Example 6 Finding the Dimensions of an American Football Field

The length of an American football field is 200 feet more than the width. If the perimeter of the field is 1040 feet, what are its dimensions?

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We know something about the length; the length is 200 feet more than the width. We will let

    x=the width.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. Because the length is 200 feet more than the width, we add 200 to the width to represent the length. Thus,

    x+200=the length.

    Figure P.16 illustrates an American football field and its dimensions.

    Figure P.16 An American football field

    An illustration shows an American football field with its width as x and length as x + 20 0.
  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. Because the perimeter of the field is 1040 feet,

    The image shows the mathematical expression 2 left parenthesis x + 20 0 right parenthesis + 2 x = 1040.

  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    Steps to solve an equation.

    Thus,

    width=x=160.length=x+200=160+200=360.

    The dimensions of an American football field are 160 feet by 360 feet. (The 360-foot length is usually described as 120 yards.)

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The perimeter of the football field using the dimensions that we found is

    2(160 feet)+2(360 feet)=320 feet+720 feet=1040 feet.

    Because the problem’s wording tells us that the perimeter is 1040 feet, our dimensions are correct.

Check Point 6

  • The length of a rectangular basketball court is 44 feet more than the width. If the perimeter of the basketball court is 288 feet, what are its dimensions?

We will use the formula for the area of a rectangle, A=lw, in our next example. The formula states that a rectangle’s area is the product of its length and its width.

Example 7 Solving a Problem Involving Landscape Design

A rectangular garden measures 80 feet by 60 feet. A large path of uniform width is to be added along both shorter sides and one longer side of the garden. The landscape designer doing the work wants to double the garden’s area with the addition of this path. How wide should the path be?

Solution

  1. Step 1 LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the width of the path.

    The situation is illustrated in Figure P.17. The figure shows the original 80-by-60 foot rectangular garden and the path of width x added along both shorter sides and one longer side.

    Figure P.17 The garden’s area is to be doubled by adding the path.

    An image shows a rectangular garden is surrounded by a path that has a length of 80 + 2 x and a width of 60 + x.
  2. Step 2 REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. Because the path is added along both shorter sides and one longer side, Figure P.17 shows that

     80+2x=the length of the new, expanded rectangle 60+x=the width of the new, expanded rectangle.

  3. Step 3 WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. The area of the rectangle must be doubled by the addition of the path.

    Left parenthesis 80 + 2 x right parenthesis, left parenthesis 60 + x right parenthesis = 2 times 80 times 60.
  4. Step 4 SOLVE THE EQUATION AND ANSWER THE QUESTION.

    (80+2x)(60+x)=28060This is the equation that modelsthe problems conditions.4800+200x+2x2=9600Multiply. Use FOIL on the left side.2x2+200x4800=0Subtract 9600 from both sidesand write the quadratic equationin standard form.2(x2+100x2400)=0Factor out 2, the GCF.2(x20)(x+120)=0Factor the trinomial.x20=0or x+120=0Set each variable factor equal to 0.x=20x=120Solve for x.

    The path cannot have a negative width. Because 120 is geometrically impossible, we use x=20. The width of the path should be 20 feet.

  5. Step 5 CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. Has the landscape architect doubled the garden’s area with the 20-foot-wide path? The area of the garden is 80 feet times 60 feet, or 4800 square feet. Because 80+2x and 60+x represent the length and width of the expanded rectangle,

    80+2x=80+220 =120 feet is the expanded rectangle's length.60+2x=60+20=80 feet is the expanded rectangle's width.

    The area of the expanded rectangle is 120 feet times 80 feet, or 9600 square feet. This is double the area of the garden, 4800 square feet, as specified by the problem’s conditions.

Check Point 7

  • A rectangular garden measures 16 feet by 12 feet. A path of uniform width is to be added so as to surround the entire garden. The landscape artist doing the work wants the garden and path to cover an area of 320 square feet. How wide should the path be?

In our next example, we will be using the Pythagorean Theorem to obtain a mathematical model. The ancient Greek philosopher and mathematician Pythagoras (approximately 582–500 b.c.) founded a school whose motto was “All is number.” Pythagoras is best remembered for his work with the right triangle, a triangle with one angle measuring 90°. The side opposite the 90° angle is called the hypotenuse. The other sides are called legs. Pythagoras found that if he constructed squares on each of the legs, as well as a larger square on the hypotenuse, the sum of the areas of the smaller squares is equal to the area of the larger square. This is illustrated in Figure P.18.

Figure P.18The area of the large square equals the sum of the areas of the smaller squares.

An illustration shows a right triangle of base 3, height 4, and hypotenuse 5.
Figure P. 18 Full Alternative Text

This relationship is usually stated in terms of the lengths of the three sides of a right triangle and is called the Pythagorean Theorem.

The Pythagorean Theorem

The sum of the squares of the lengths of the legs of a right triangle equals the square of the length of the hypotenuse.

If the legs have lengths a and b, and the hypotenuse has length c, then

a2+b2=c2.
An image shows a right angle triangle ABC, which represents angle C is a 90 degree angle. AC represents as a base labeled, b, Leg, BC denotes as a height labeled, a Leg, and AB is labeled, c, hypotenuse.

Example 8 Using the Pythagorean Theorem: Screen Math

A bar graph shows the rising percentage of large screen TVs on store shelves.

Source: GAP Intelligence

Did you know that the size of a television screen refers to the length of its diagonal? If the length of the HDTV screen in Figure P.19 is 28 inches and its width is 15.7 inches, what is the size of the screen to the nearest inch?

Figure P.19

A photo shows a television with a screen size of length 28 inches and a width of 15.7 inches.

Solution

Figure P.19 shows that the length, width, and diagonal of the screen form a right triangle. The diagonal is the hypotenuse of the triangle. We use the Pythagorean Theorem with a=28, b=15.7, and solve for the screen size, c.

a2+b2=c2This is the Pythagorean Theorem.282+15.72=c2Let a=28 and b=15.7.784+246.49=c2282=2828=784 and15.72=15.715.7=246.49.c2=1030.49Add: 784+246.49=1030.49. We also reversedthe two sides.
c=1030.49 or  c=1030.49Apply the square root property.c32 or  c32Use a calculator and round to thenearest inch. 1030.49    = or   1030.49 ENTER

Because c represents the size of the screen, this dimension must be positive. We reject 32. Thus, the screen size of the HDTV is 32 inches.

Check Point 8

  • Figure P.20 shows the dimensions of an old TV screen. What is the size of the screen?

    Figure P.20

    A photo shows a television with a screen size of length 25.6 inches and a width of 19.2 inches.

Example 9 Dividing the Cost of a Yacht

A group of friends agrees to share the cost of a $50,000 yacht equally. Before the purchase is made, one more person joins the group. As a result, each person’s share is reduced by $2500. How many people were in the original group?

Solution

  1. Step 1 LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the number of people in the original group.
  2. Step 2 REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x.  Because one more person joined the original group, let

    x+1=the number of people in the final group.
  3. Step 3 WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. As a result of one more person joining the original group, each person’s share is reduced by $2500.

    The image shows a mathematical expression, start fraction 50,000 over x end fraction minus 2500 = start fraction 50,000 over x + 1 end fraction.
  4. Step 4 SOLVE THE EQUATION AND ANSWER THE QUESTION.

    50,000x2500=50,000x+1This is the equation that models theproblems conditions.x(x+1)(50,000x2500)=x(x+1)50,000x+1Multiply both sides by x(x+1),the LCD. x (x+1)50,000 x x(x+1)2500=x (x+1) 50,000 (x+1) Use the distributive property anddivide out common factors.50,000(x+1)2500x(x+1)=50,000xSimplify.50,000x+50,0002500x22500x=50,000xUse the distributive property.2500x2+47,500x+50,000=50,000xCombine like terms.2500x22500x+50,000=0Write the quadratic equation instandard form.2500(x2+x20)=0Factor out 2500.2500(x+5)(x4)=0Factor completely.x+5=0or x4=0Set each variable factor equal to zero.x=5x=4Solve the resulting equations.

    Because x represents the number of people in the original group, x cannot be negative. Thus, there were four people in the original group.

  5. Step 5 CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM.

     original cost per person=$50,0004=$12,500 final cost per person=$50,0005=$10,000

    We see that each person’s share is reduced by $12,500$10,000, or $2500, as specified by the problem’s conditions.

Check Point 9

  • A group of people share equally in a $5,000,000 lottery. Before the money is divided, three more winning ticket holders are declared. As a result, each person’s share is reduced by $375,000. How many people were in the original group of winners?

Objective 1: Use equations to solve problems

Problem Solving with Equations

  1. Objective 1Use equations to solve problems.

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We have seen that a model is a mathematical representation of a real-world situation. In this section, we will be solving problems that are presented in English. This means that we must obtain models by translating from the ordinary language of English into the language of algebraic equations. To translate, however, we must understand the English prose and be familiar with the forms of algebraic language. Following are some general steps we will use in solving word problems:

Strategy for Solving Word Problems

  1. Step 1 Read the problem carefully several times until you can state in your own words what is given and what the problem is looking for. Let x (or any variable) represent one of the unknown quantities in the problem.

  2. Step 2 If necessary, write expressions for any other unknown quantities in the problem in terms of x.

  3. Step 3 Write an equation in x that models the verbal conditions of the problem.

  4. Step 4 Solve the equation and answer the problem’s question.

  5. Step 5 Check the solution in the original wording of the problem, not in the equation obtained from the words.

Example 1 Education Pays Off

The graph in Figure P.14 shows median yearly earnings in the United States by highest educational attainment. It certainly looks like you can make more money with more education. However, workers who make more money pay a higher percentage of income in taxes, so it is wise to consider after-tax earnings when deciding if that higher degree will pay off.

Figure P.14

A bar chart shows median earnings of full-time workers in the United States, by highest educational attainment.

Source: Education Pays 2019

The median yearly after-tax income of a full-time worker with a bachelor’s degree exceeds that of a full-time worker with an associate’s degree by $11 thousand. The median yearly after-tax income of a full-time worker with a master’s degree exceeds that of a full-time worker with an associate’s degree by $21 thousand. Combined, three full-time workers with each of these degrees earn $149 thousand after taxes. Find the median yearly after-tax income of full-time workers with each of these levels of education.

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We know something about after-tax incomes of full-time workers with bachelor’s degrees and master’s degrees: They exceed the after-tax income of a full-time worker with an associate’s degree by $11 thousand and $21 thousand, respectively. We will let

    x=the median yearly after-tax income of a full-time worker with an associates degree.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. Because a full-time worker with a bachelor’s degree earns $11 thousand more after taxes than a full-time worker with an associate’s degree, let

    x+11=the median yearly after-tax income of a full-time worker with a bachelors degree.

    Because a full-time worker with a master’s degree earns $21 thousand more after taxes than a full-time worker with an associate’s degree, let

    x+21=the median yearly after-tax income of a full-time worker with a master's degree.
  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. Combined, three full-time workers with each of these degrees earn $149 thousand.

    An equation with variable x to model the conditions.

  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    x+(x+11)+(x+21)=149This is the equation that models the problems conditions.3x+32=149Remove parentheses, regroup, and combine like terms.3x=117Subtract 32 from both sides.x=39Divide both sides by 3.

    Because we isolated the variable in the model and obtained x=39,

    median after-tax income with an associates degree=x=39median after-tax income with a bachelors degree=x+11=39+11=50median after-tax income with a masters degree=x+21=39+21=60.

    Full-time workers with associate’s degrees earn $39 thousand per year after taxes, full-time workers with bachelor’s degrees earn $50 thousand per year after taxes, and full-time workers with master’s degrees earn $60 thousand per year after taxes.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that combined, three full-time workers with each of these educational attainments earn $149 thousand after taxes. Using the incomes we determined in step 4, the sum is

    $39 thousand+$50 thousand+$60 thousand,or $149 thousand,

    which satisfies the problem’s conditions.

Check Point 1

  • The median yearly before-tax income of a full-time worker with a bachelor’s degree exceeds that of a full-time worker with an associate’s degree by $15 thousand. The median yearly before-tax income of a full-time worker with a master’s degree exceeds that of a full-time worker with an associate’s degree by $30 thousand. Combined, three full-time workers with each of these educational attainments earn $195 thousand before taxes. Find the median yearly before-tax income of full-time workers with each of these levels of education. (These incomes are illustrated by the bar graph on the previous page.)

Example 2 Modeling Attitudes of College Freshmen

Your author teaching math in 1969

Researchers have surveyed college freshmen every year since 1969. Figure P.15 shows that attitudes about some life goals have changed dramatically over the years. In particular, the freshman class of 2018 was more interested in making money than the freshmen of 1969 had been. In 1969, 42% of first-year college students considered “being well-off financially” essential or very important. For the period from 1969 through 2018, this percentage increased by approximately 0.8 each year. If this trend continues, by which year will all college freshmen consider “being well-off financially” essential or very important?

Figure P.15

A bar chart shows the life objectives of college freshmen, for the period from 19 69 to 20 18.

Source: Higher Education Research Institute

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We are interested in the year when all college freshmen, or 100% of the freshmen, will consider this life objective essential or very important. Let

    x = the number of years after 1969 when all freshmen will considerbeing    well-off financially essential or very important.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. There are no other unknown quantities to find, so we can skip this step.

  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS.

    The image shows a mathematical expression 42 + 0.8 x = 100
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    42+0.8x=100This is the equation that models the problems conditions.4242+0.8x=10042Subtract 42 from both sides.0.8x=58Simplify.0.8x0.8=580.8Divide both sides by 0.8.x=72.573Simplify and round to the nearestwhole number.

    Using current trends, by approximately 73 years after 1969, or in 2042, all freshmen will consider “being well-off financially” essential or very important.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that all freshmen (100%, represented by 100 using the model) will consider the objective essential or very important. Does this approximately occur if we increase the 1969 percentage, 42, by 0.8 each year for 73 years, our proposed solution?

    42+0.8(73)=42+58.4=100.4100

    This verifies that using trends shown in Figure P.15, all first-year college students will consider the objective of being well-off financially essential or very important approximately 73 years after 1969, or in 2042.

Check Point 2

  • Figure P.15 on the previous page shows that the freshman class of 2018 was less interested in developing a philosophy of life than the freshmen of 1969 had been. In 1969, 85% of the freshmen considered this objective essential or very important. Since then, this percentage has decreased by approximately 0.8 each year. If this trend continues, by which year will only 25% of college freshmen consider “developing a meaningful philosophy of life” essential or very important?

Example 3 Modeling Options for a Toll

It’s time to decide how you’re going to pay the bridge toll so that you can get to the beach. The first option requires purchasing a transponder for $20; with the transponder, you pay a reduced toll of $3.25 each time you cross the bridge. Your second option is toll-by-plate; with this option, you pay the full toll of $4.25 each time you cross the bridge plus a $3 administrative fee for each crossing. Find the number of times you would need to cross the bridge for the costs of the two options to be the same.

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. Let

    x=the number of times you cross the bridge.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. There are no other unknown quantities, so we can skip this step.

  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. The cost with the transponder is the cost of the transponder, $20, plus the toll, $3.25, times the number of crossings, x. For toll-by-plate, you must pay both the toll, $4.25, and the administrative fee, $3, each time you cross the bridge. The cost for toll-by-plate is the sum of the toll and the administrative fee, $4.25+$3 or $7.25, times the number of crossings, x. We want to know when these two costs are equal.

    The image shows a mathematical expression 7.25x = 20 + 3.25 x
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    7.25x=20+3.25xThis is the equation that models theproblems conditions.4x=20Subtract 3.25x from both sides.4x=5Divide both sides by 4.

    Because x represents the number of times you cross the bridge, the total cost of toll-by-plate is the same as the total cost with the transponder for 5 crossings.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that the cost of toll-by-plate should be the same as the cost with the transponder. Let’s see if they are the same with 5 bridge crossings per month.

    Expression to check the proposed solution.

    If you cross the bridge five times, both options cost $36.25. Thus the proposed solution, 5 crossings, satisfies the problem’s conditions.

Check Point 3

  • You drive up to a toll plaza and find booths with attendants where you can pay the toll by cash or credit card. With this option, the toll is $5 each time you cross the bridge. The attendant gives you the option of buying a bar-coded decal for $25; with the decal, you get 25% off the normal toll of $5 for each crossing. Find the number of times you would need to cross the bridge for the costs of the two options to be the same.

Example 4 A Price Reduction on Wireless Headphones

Your favorite electronics retailer is having a terrific sale on noise cancelling wireless headphones. (You’ll finally be able to listen to music in peace!) After a 40% price reduction, you purchase headphones for $192. What was the price of the headphones before the reduction?

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the price of the headphones prior to the reduction.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. There are no other unknown quantities to find, so we can skip this step.

  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. The headphones’ original price minus the 40% reduction is the reduced price, $192.

    The image shows a mathematical expression x minus 0.4 x = 192.
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    The solution of an equation.

    The headphones’ price before the reduction was $320.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The price before the reduction, $320, minus the 40% reduction should equal the reduced price given in the original wording, $192:

    32040% of 320=3200.4(320)=320128=192.

    This verifies that the headphones’ price before the reduction was $320.

Check Point 4

  • After a 30% price reduction, you purchase a new laptop for $840. What was the laptop’s price before the reduction?

Our next example is about simple interest. Simple interest involves interest calculated only on the amount of money that we invest, called the principal. The formula I=Pr is used to find the simple interest, I, earned for one year when the principal, P, is invested at an annual interest rate, r. Dual investment problems involve different amounts of money in two or more investments, each paying a different rate.

Example 5 Solving a Dual Investment Problem

Your grandmother needs your help. She has $150,000 to invest. Part of this money is to be invested in a money market account paying 1.5% annual interest. The rest of this money is to be invested in a certificate of deposit paying 1.3% annual interest. She told you that she requires $2000 per year in extra income from the combination of these investments. How much money should be placed in each investment?

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the amount invested in the money market account at 1.5%.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. The other quantity that we seek is the amount invested at 1.3% in the certificate of deposit. Because the total amount Grandma has to invest is $150,000 and we already used up x,

    150,000x=the amount invested in the certificate of deposit at 1.3%.
  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. Because Grandma requires $2000 in total interest, the interest for the two investments combined must be $2000. Interest is Pr or rP for each investment.

    The image shows a mathematical expression 0.015 x + 0.013 left parenthesis 150,000 minus x right parenthesis = 20 00
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    A mathematical expression steps to solve an equation to find the value of x.

    Thus,

    the amount invested at1.5%=x=25,000the amount invested at 1.3%=150,00025,000=125,000

    Grandma should invest $25,000 at 1.5% and $125,000 at 1.3%.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that the total interest from the dual investments should be $2000. Can Grandma count on $2000 interest? The interest earned on $25,000 at 1.5% is ($25,000) (0.015), or $375. The interest earned on $125,000 at 1.3% is ($125,000)(0.013), or $1625. The total interest is $375+$1625, or $2000, exactly as it should be. You’ve made your grandmother happy. (Now if you would just visit her more often …)

Check Point 5

  • You inherited $50,000 with the stipulation that for the first year the money had to be invested in two accounts paying 0.9% and 1.1% annual interest. How much did you invest at each rate if the total interest earned for the year was $515?

Solving geometry problems usually requires a knowledge of basic geometric ideas and formulas. Formulas for area, perimeter, and volume are given in Table P.6.

Table P.6 Common Formulas for Area, Perimeter, and Volume

An illustration shows common formulas for area, perimeter, and volume of different shapes.
Table P.6 Full Alternative Text

We will be using the formula for the perimeter of a rectangle, P=2l+2w, in our next example. The formula states that a rectangle’s perimeter is the sum of twice its length and twice its width.

Example 6 Finding the Dimensions of an American Football Field

The length of an American football field is 200 feet more than the width. If the perimeter of the field is 1040 feet, what are its dimensions?

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We know something about the length; the length is 200 feet more than the width. We will let

    x=the width.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. Because the length is 200 feet more than the width, we add 200 to the width to represent the length. Thus,

    x+200=the length.

    Figure P.16 illustrates an American football field and its dimensions.

    Figure P.16 An American football field

    An illustration shows an American football field with its width as x and length as x + 20 0.
  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. Because the perimeter of the field is 1040 feet,

    The image shows the mathematical expression 2 left parenthesis x + 20 0 right parenthesis + 2 x = 1040.

  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    Steps to solve an equation.

    Thus,

    width=x=160.length=x+200=160+200=360.

    The dimensions of an American football field are 160 feet by 360 feet. (The 360-foot length is usually described as 120 yards.)

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The perimeter of the football field using the dimensions that we found is

    2(160 feet)+2(360 feet)=320 feet+720 feet=1040 feet.

    Because the problem’s wording tells us that the perimeter is 1040 feet, our dimensions are correct.

Check Point 6

  • The length of a rectangular basketball court is 44 feet more than the width. If the perimeter of the basketball court is 288 feet, what are its dimensions?

We will use the formula for the area of a rectangle, A=lw, in our next example. The formula states that a rectangle’s area is the product of its length and its width.

Example 7 Solving a Problem Involving Landscape Design

A rectangular garden measures 80 feet by 60 feet. A large path of uniform width is to be added along both shorter sides and one longer side of the garden. The landscape designer doing the work wants to double the garden’s area with the addition of this path. How wide should the path be?

Solution

  1. Step 1 LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the width of the path.

    The situation is illustrated in Figure P.17. The figure shows the original 80-by-60 foot rectangular garden and the path of width x added along both shorter sides and one longer side.

    Figure P.17 The garden’s area is to be doubled by adding the path.

    An image shows a rectangular garden is surrounded by a path that has a length of 80 + 2 x and a width of 60 + x.
  2. Step 2 REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. Because the path is added along both shorter sides and one longer side, Figure P.17 shows that

     80+2x=the length of the new, expanded rectangle 60+x=the width of the new, expanded rectangle.

  3. Step 3 WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. The area of the rectangle must be doubled by the addition of the path.

    Left parenthesis 80 + 2 x right parenthesis, left parenthesis 60 + x right parenthesis = 2 times 80 times 60.
  4. Step 4 SOLVE THE EQUATION AND ANSWER THE QUESTION.

    (80+2x)(60+x)=28060This is the equation that modelsthe problems conditions.4800+200x+2x2=9600Multiply. Use FOIL on the left side.2x2+200x4800=0Subtract 9600 from both sidesand write the quadratic equationin standard form.2(x2+100x2400)=0Factor out 2, the GCF.2(x20)(x+120)=0Factor the trinomial.x20=0or x+120=0Set each variable factor equal to 0.x=20x=120Solve for x.

    The path cannot have a negative width. Because 120 is geometrically impossible, we use x=20. The width of the path should be 20 feet.

  5. Step 5 CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. Has the landscape architect doubled the garden’s area with the 20-foot-wide path? The area of the garden is 80 feet times 60 feet, or 4800 square feet. Because 80+2x and 60+x represent the length and width of the expanded rectangle,

    80+2x=80+220 =120 feet is the expanded rectangle's length.60+2x=60+20=80 feet is the expanded rectangle's width.

    The area of the expanded rectangle is 120 feet times 80 feet, or 9600 square feet. This is double the area of the garden, 4800 square feet, as specified by the problem’s conditions.

Check Point 7

  • A rectangular garden measures 16 feet by 12 feet. A path of uniform width is to be added so as to surround the entire garden. The landscape artist doing the work wants the garden and path to cover an area of 320 square feet. How wide should the path be?

In our next example, we will be using the Pythagorean Theorem to obtain a mathematical model. The ancient Greek philosopher and mathematician Pythagoras (approximately 582–500 b.c.) founded a school whose motto was “All is number.” Pythagoras is best remembered for his work with the right triangle, a triangle with one angle measuring 90°. The side opposite the 90° angle is called the hypotenuse. The other sides are called legs. Pythagoras found that if he constructed squares on each of the legs, as well as a larger square on the hypotenuse, the sum of the areas of the smaller squares is equal to the area of the larger square. This is illustrated in Figure P.18.

Figure P.18The area of the large square equals the sum of the areas of the smaller squares.

An illustration shows a right triangle of base 3, height 4, and hypotenuse 5.
Figure P. 18 Full Alternative Text

This relationship is usually stated in terms of the lengths of the three sides of a right triangle and is called the Pythagorean Theorem.

The Pythagorean Theorem

The sum of the squares of the lengths of the legs of a right triangle equals the square of the length of the hypotenuse.

If the legs have lengths a and b, and the hypotenuse has length c, then

a2+b2=c2.
An image shows a right angle triangle ABC, which represents angle C is a 90 degree angle. AC represents as a base labeled, b, Leg, BC denotes as a height labeled, a Leg, and AB is labeled, c, hypotenuse.

Example 8 Using the Pythagorean Theorem: Screen Math

A bar graph shows the rising percentage of large screen TVs on store shelves.

Source: GAP Intelligence

Did you know that the size of a television screen refers to the length of its diagonal? If the length of the HDTV screen in Figure P.19 is 28 inches and its width is 15.7 inches, what is the size of the screen to the nearest inch?

Figure P.19

A photo shows a television with a screen size of length 28 inches and a width of 15.7 inches.

Solution

Figure P.19 shows that the length, width, and diagonal of the screen form a right triangle. The diagonal is the hypotenuse of the triangle. We use the Pythagorean Theorem with a=28, b=15.7, and solve for the screen size, c.

a2+b2=c2This is the Pythagorean Theorem.282+15.72=c2Let a=28 and b=15.7.784+246.49=c2282=2828=784 and15.72=15.715.7=246.49.c2=1030.49Add: 784+246.49=1030.49. We also reversedthe two sides.
c=1030.49 or  c=1030.49Apply the square root property.c32 or  c32Use a calculator and round to thenearest inch. 1030.49    = or   1030.49 ENTER

Because c represents the size of the screen, this dimension must be positive. We reject 32. Thus, the screen size of the HDTV is 32 inches.

Check Point 8

  • Figure P.20 shows the dimensions of an old TV screen. What is the size of the screen?

    Figure P.20

    A photo shows a television with a screen size of length 25.6 inches and a width of 19.2 inches.

Example 9 Dividing the Cost of a Yacht

A group of friends agrees to share the cost of a $50,000 yacht equally. Before the purchase is made, one more person joins the group. As a result, each person’s share is reduced by $2500. How many people were in the original group?

Solution

  1. Step 1 LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the number of people in the original group.
  2. Step 2 REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x.  Because one more person joined the original group, let

    x+1=the number of people in the final group.
  3. Step 3 WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. As a result of one more person joining the original group, each person’s share is reduced by $2500.

    The image shows a mathematical expression, start fraction 50,000 over x end fraction minus 2500 = start fraction 50,000 over x + 1 end fraction.
  4. Step 4 SOLVE THE EQUATION AND ANSWER THE QUESTION.

    50,000x2500=50,000x+1This is the equation that models theproblems conditions.x(x+1)(50,000x2500)=x(x+1)50,000x+1Multiply both sides by x(x+1),the LCD. x (x+1)50,000 x x(x+1)2500=x (x+1) 50,000 (x+1) Use the distributive property anddivide out common factors.50,000(x+1)2500x(x+1)=50,000xSimplify.50,000x+50,0002500x22500x=50,000xUse the distributive property.2500x2+47,500x+50,000=50,000xCombine like terms.2500x22500x+50,000=0Write the quadratic equation instandard form.2500(x2+x20)=0Factor out 2500.2500(x+5)(x4)=0Factor completely.x+5=0or x4=0Set each variable factor equal to zero.x=5x=4Solve the resulting equations.

    Because x represents the number of people in the original group, x cannot be negative. Thus, there were four people in the original group.

  5. Step 5 CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM.

     original cost per person=$50,0004=$12,500 final cost per person=$50,0005=$10,000

    We see that each person’s share is reduced by $12,500$10,000, or $2500, as specified by the problem’s conditions.

Check Point 9

  • A group of people share equally in a $5,000,000 lottery. Before the money is divided, three more winning ticket holders are declared. As a result, each person’s share is reduced by $375,000. How many people were in the original group of winners?

Objective 1: Use equations to solve problems

Problem Solving with Equations

  1. Objective 1Use equations to solve problems.

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We have seen that a model is a mathematical representation of a real-world situation. In this section, we will be solving problems that are presented in English. This means that we must obtain models by translating from the ordinary language of English into the language of algebraic equations. To translate, however, we must understand the English prose and be familiar with the forms of algebraic language. Following are some general steps we will use in solving word problems:

Strategy for Solving Word Problems

  1. Step 1 Read the problem carefully several times until you can state in your own words what is given and what the problem is looking for. Let x (or any variable) represent one of the unknown quantities in the problem.

  2. Step 2 If necessary, write expressions for any other unknown quantities in the problem in terms of x.

  3. Step 3 Write an equation in x that models the verbal conditions of the problem.

  4. Step 4 Solve the equation and answer the problem’s question.

  5. Step 5 Check the solution in the original wording of the problem, not in the equation obtained from the words.

Example 1 Education Pays Off

The graph in Figure P.14 shows median yearly earnings in the United States by highest educational attainment. It certainly looks like you can make more money with more education. However, workers who make more money pay a higher percentage of income in taxes, so it is wise to consider after-tax earnings when deciding if that higher degree will pay off.

Figure P.14

A bar chart shows median earnings of full-time workers in the United States, by highest educational attainment.

Source: Education Pays 2019

The median yearly after-tax income of a full-time worker with a bachelor’s degree exceeds that of a full-time worker with an associate’s degree by $11 thousand. The median yearly after-tax income of a full-time worker with a master’s degree exceeds that of a full-time worker with an associate’s degree by $21 thousand. Combined, three full-time workers with each of these degrees earn $149 thousand after taxes. Find the median yearly after-tax income of full-time workers with each of these levels of education.

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We know something about after-tax incomes of full-time workers with bachelor’s degrees and master’s degrees: They exceed the after-tax income of a full-time worker with an associate’s degree by $11 thousand and $21 thousand, respectively. We will let

    x=the median yearly after-tax income of a full-time worker with an associates degree.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. Because a full-time worker with a bachelor’s degree earns $11 thousand more after taxes than a full-time worker with an associate’s degree, let

    x+11=the median yearly after-tax income of a full-time worker with a bachelors degree.

    Because a full-time worker with a master’s degree earns $21 thousand more after taxes than a full-time worker with an associate’s degree, let

    x+21=the median yearly after-tax income of a full-time worker with a master's degree.
  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. Combined, three full-time workers with each of these degrees earn $149 thousand.

    An equation with variable x to model the conditions.

  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    x+(x+11)+(x+21)=149This is the equation that models the problems conditions.3x+32=149Remove parentheses, regroup, and combine like terms.3x=117Subtract 32 from both sides.x=39Divide both sides by 3.

    Because we isolated the variable in the model and obtained x=39,

    median after-tax income with an associates degree=x=39median after-tax income with a bachelors degree=x+11=39+11=50median after-tax income with a masters degree=x+21=39+21=60.

    Full-time workers with associate’s degrees earn $39 thousand per year after taxes, full-time workers with bachelor’s degrees earn $50 thousand per year after taxes, and full-time workers with master’s degrees earn $60 thousand per year after taxes.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that combined, three full-time workers with each of these educational attainments earn $149 thousand after taxes. Using the incomes we determined in step 4, the sum is

    $39 thousand+$50 thousand+$60 thousand,or $149 thousand,

    which satisfies the problem’s conditions.

Check Point 1

  • The median yearly before-tax income of a full-time worker with a bachelor’s degree exceeds that of a full-time worker with an associate’s degree by $15 thousand. The median yearly before-tax income of a full-time worker with a master’s degree exceeds that of a full-time worker with an associate’s degree by $30 thousand. Combined, three full-time workers with each of these educational attainments earn $195 thousand before taxes. Find the median yearly before-tax income of full-time workers with each of these levels of education. (These incomes are illustrated by the bar graph on the previous page.)

Example 2 Modeling Attitudes of College Freshmen

Your author teaching math in 1969

Researchers have surveyed college freshmen every year since 1969. Figure P.15 shows that attitudes about some life goals have changed dramatically over the years. In particular, the freshman class of 2018 was more interested in making money than the freshmen of 1969 had been. In 1969, 42% of first-year college students considered “being well-off financially” essential or very important. For the period from 1969 through 2018, this percentage increased by approximately 0.8 each year. If this trend continues, by which year will all college freshmen consider “being well-off financially” essential or very important?

Figure P.15

A bar chart shows the life objectives of college freshmen, for the period from 19 69 to 20 18.

Source: Higher Education Research Institute

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We are interested in the year when all college freshmen, or 100% of the freshmen, will consider this life objective essential or very important. Let

    x = the number of years after 1969 when all freshmen will considerbeing    well-off financially essential or very important.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. There are no other unknown quantities to find, so we can skip this step.

  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS.

    The image shows a mathematical expression 42 + 0.8 x = 100
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    42+0.8x=100This is the equation that models the problems conditions.4242+0.8x=10042Subtract 42 from both sides.0.8x=58Simplify.0.8x0.8=580.8Divide both sides by 0.8.x=72.573Simplify and round to the nearestwhole number.

    Using current trends, by approximately 73 years after 1969, or in 2042, all freshmen will consider “being well-off financially” essential or very important.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that all freshmen (100%, represented by 100 using the model) will consider the objective essential or very important. Does this approximately occur if we increase the 1969 percentage, 42, by 0.8 each year for 73 years, our proposed solution?

    42+0.8(73)=42+58.4=100.4100

    This verifies that using trends shown in Figure P.15, all first-year college students will consider the objective of being well-off financially essential or very important approximately 73 years after 1969, or in 2042.

Check Point 2

  • Figure P.15 on the previous page shows that the freshman class of 2018 was less interested in developing a philosophy of life than the freshmen of 1969 had been. In 1969, 85% of the freshmen considered this objective essential or very important. Since then, this percentage has decreased by approximately 0.8 each year. If this trend continues, by which year will only 25% of college freshmen consider “developing a meaningful philosophy of life” essential or very important?

Example 3 Modeling Options for a Toll

It’s time to decide how you’re going to pay the bridge toll so that you can get to the beach. The first option requires purchasing a transponder for $20; with the transponder, you pay a reduced toll of $3.25 each time you cross the bridge. Your second option is toll-by-plate; with this option, you pay the full toll of $4.25 each time you cross the bridge plus a $3 administrative fee for each crossing. Find the number of times you would need to cross the bridge for the costs of the two options to be the same.

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. Let

    x=the number of times you cross the bridge.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. There are no other unknown quantities, so we can skip this step.

  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. The cost with the transponder is the cost of the transponder, $20, plus the toll, $3.25, times the number of crossings, x. For toll-by-plate, you must pay both the toll, $4.25, and the administrative fee, $3, each time you cross the bridge. The cost for toll-by-plate is the sum of the toll and the administrative fee, $4.25+$3 or $7.25, times the number of crossings, x. We want to know when these two costs are equal.

    The image shows a mathematical expression 7.25x = 20 + 3.25 x
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    7.25x=20+3.25xThis is the equation that models theproblems conditions.4x=20Subtract 3.25x from both sides.4x=5Divide both sides by 4.

    Because x represents the number of times you cross the bridge, the total cost of toll-by-plate is the same as the total cost with the transponder for 5 crossings.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that the cost of toll-by-plate should be the same as the cost with the transponder. Let’s see if they are the same with 5 bridge crossings per month.

    Expression to check the proposed solution.

    If you cross the bridge five times, both options cost $36.25. Thus the proposed solution, 5 crossings, satisfies the problem’s conditions.

Check Point 3

  • You drive up to a toll plaza and find booths with attendants where you can pay the toll by cash or credit card. With this option, the toll is $5 each time you cross the bridge. The attendant gives you the option of buying a bar-coded decal for $25; with the decal, you get 25% off the normal toll of $5 for each crossing. Find the number of times you would need to cross the bridge for the costs of the two options to be the same.

Example 4 A Price Reduction on Wireless Headphones

Your favorite electronics retailer is having a terrific sale on noise cancelling wireless headphones. (You’ll finally be able to listen to music in peace!) After a 40% price reduction, you purchase headphones for $192. What was the price of the headphones before the reduction?

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the price of the headphones prior to the reduction.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. There are no other unknown quantities to find, so we can skip this step.

  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. The headphones’ original price minus the 40% reduction is the reduced price, $192.

    The image shows a mathematical expression x minus 0.4 x = 192.
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    The solution of an equation.

    The headphones’ price before the reduction was $320.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The price before the reduction, $320, minus the 40% reduction should equal the reduced price given in the original wording, $192:

    32040% of 320=3200.4(320)=320128=192.

    This verifies that the headphones’ price before the reduction was $320.

Check Point 4

  • After a 30% price reduction, you purchase a new laptop for $840. What was the laptop’s price before the reduction?

Our next example is about simple interest. Simple interest involves interest calculated only on the amount of money that we invest, called the principal. The formula I=Pr is used to find the simple interest, I, earned for one year when the principal, P, is invested at an annual interest rate, r. Dual investment problems involve different amounts of money in two or more investments, each paying a different rate.

Example 5 Solving a Dual Investment Problem

Your grandmother needs your help. She has $150,000 to invest. Part of this money is to be invested in a money market account paying 1.5% annual interest. The rest of this money is to be invested in a certificate of deposit paying 1.3% annual interest. She told you that she requires $2000 per year in extra income from the combination of these investments. How much money should be placed in each investment?

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the amount invested in the money market account at 1.5%.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. The other quantity that we seek is the amount invested at 1.3% in the certificate of deposit. Because the total amount Grandma has to invest is $150,000 and we already used up x,

    150,000x=the amount invested in the certificate of deposit at 1.3%.
  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. Because Grandma requires $2000 in total interest, the interest for the two investments combined must be $2000. Interest is Pr or rP for each investment.

    The image shows a mathematical expression 0.015 x + 0.013 left parenthesis 150,000 minus x right parenthesis = 20 00
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    A mathematical expression steps to solve an equation to find the value of x.

    Thus,

    the amount invested at1.5%=x=25,000the amount invested at 1.3%=150,00025,000=125,000

    Grandma should invest $25,000 at 1.5% and $125,000 at 1.3%.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that the total interest from the dual investments should be $2000. Can Grandma count on $2000 interest? The interest earned on $25,000 at 1.5% is ($25,000) (0.015), or $375. The interest earned on $125,000 at 1.3% is ($125,000)(0.013), or $1625. The total interest is $375+$1625, or $2000, exactly as it should be. You’ve made your grandmother happy. (Now if you would just visit her more often …)

Check Point 5

  • You inherited $50,000 with the stipulation that for the first year the money had to be invested in two accounts paying 0.9% and 1.1% annual interest. How much did you invest at each rate if the total interest earned for the year was $515?

Solving geometry problems usually requires a knowledge of basic geometric ideas and formulas. Formulas for area, perimeter, and volume are given in Table P.6.

Table P.6 Common Formulas for Area, Perimeter, and Volume

An illustration shows common formulas for area, perimeter, and volume of different shapes.
Table P.6 Full Alternative Text

We will be using the formula for the perimeter of a rectangle, P=2l+2w, in our next example. The formula states that a rectangle’s perimeter is the sum of twice its length and twice its width.

Example 6 Finding the Dimensions of an American Football Field

The length of an American football field is 200 feet more than the width. If the perimeter of the field is 1040 feet, what are its dimensions?

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We know something about the length; the length is 200 feet more than the width. We will let

    x=the width.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. Because the length is 200 feet more than the width, we add 200 to the width to represent the length. Thus,

    x+200=the length.

    Figure P.16 illustrates an American football field and its dimensions.

    Figure P.16 An American football field

    An illustration shows an American football field with its width as x and length as x + 20 0.
  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. Because the perimeter of the field is 1040 feet,

    The image shows the mathematical expression 2 left parenthesis x + 20 0 right parenthesis + 2 x = 1040.

  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    Steps to solve an equation.

    Thus,

    width=x=160.length=x+200=160+200=360.

    The dimensions of an American football field are 160 feet by 360 feet. (The 360-foot length is usually described as 120 yards.)

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The perimeter of the football field using the dimensions that we found is

    2(160 feet)+2(360 feet)=320 feet+720 feet=1040 feet.

    Because the problem’s wording tells us that the perimeter is 1040 feet, our dimensions are correct.

Check Point 6

  • The length of a rectangular basketball court is 44 feet more than the width. If the perimeter of the basketball court is 288 feet, what are its dimensions?

We will use the formula for the area of a rectangle, A=lw, in our next example. The formula states that a rectangle’s area is the product of its length and its width.

Example 7 Solving a Problem Involving Landscape Design

A rectangular garden measures 80 feet by 60 feet. A large path of uniform width is to be added along both shorter sides and one longer side of the garden. The landscape designer doing the work wants to double the garden’s area with the addition of this path. How wide should the path be?

Solution

  1. Step 1 LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the width of the path.

    The situation is illustrated in Figure P.17. The figure shows the original 80-by-60 foot rectangular garden and the path of width x added along both shorter sides and one longer side.

    Figure P.17 The garden’s area is to be doubled by adding the path.

    An image shows a rectangular garden is surrounded by a path that has a length of 80 + 2 x and a width of 60 + x.
  2. Step 2 REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. Because the path is added along both shorter sides and one longer side, Figure P.17 shows that

     80+2x=the length of the new, expanded rectangle 60+x=the width of the new, expanded rectangle.

  3. Step 3 WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. The area of the rectangle must be doubled by the addition of the path.

    Left parenthesis 80 + 2 x right parenthesis, left parenthesis 60 + x right parenthesis = 2 times 80 times 60.
  4. Step 4 SOLVE THE EQUATION AND ANSWER THE QUESTION.

    (80+2x)(60+x)=28060This is the equation that modelsthe problems conditions.4800+200x+2x2=9600Multiply. Use FOIL on the left side.2x2+200x4800=0Subtract 9600 from both sidesand write the quadratic equationin standard form.2(x2+100x2400)=0Factor out 2, the GCF.2(x20)(x+120)=0Factor the trinomial.x20=0or x+120=0Set each variable factor equal to 0.x=20x=120Solve for x.

    The path cannot have a negative width. Because 120 is geometrically impossible, we use x=20. The width of the path should be 20 feet.

  5. Step 5 CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. Has the landscape architect doubled the garden’s area with the 20-foot-wide path? The area of the garden is 80 feet times 60 feet, or 4800 square feet. Because 80+2x and 60+x represent the length and width of the expanded rectangle,

    80+2x=80+220 =120 feet is the expanded rectangle's length.60+2x=60+20=80 feet is the expanded rectangle's width.

    The area of the expanded rectangle is 120 feet times 80 feet, or 9600 square feet. This is double the area of the garden, 4800 square feet, as specified by the problem’s conditions.

Check Point 7

  • A rectangular garden measures 16 feet by 12 feet. A path of uniform width is to be added so as to surround the entire garden. The landscape artist doing the work wants the garden and path to cover an area of 320 square feet. How wide should the path be?

In our next example, we will be using the Pythagorean Theorem to obtain a mathematical model. The ancient Greek philosopher and mathematician Pythagoras (approximately 582–500 b.c.) founded a school whose motto was “All is number.” Pythagoras is best remembered for his work with the right triangle, a triangle with one angle measuring 90°. The side opposite the 90° angle is called the hypotenuse. The other sides are called legs. Pythagoras found that if he constructed squares on each of the legs, as well as a larger square on the hypotenuse, the sum of the areas of the smaller squares is equal to the area of the larger square. This is illustrated in Figure P.18.

Figure P.18The area of the large square equals the sum of the areas of the smaller squares.

An illustration shows a right triangle of base 3, height 4, and hypotenuse 5.
Figure P. 18 Full Alternative Text

This relationship is usually stated in terms of the lengths of the three sides of a right triangle and is called the Pythagorean Theorem.

The Pythagorean Theorem

The sum of the squares of the lengths of the legs of a right triangle equals the square of the length of the hypotenuse.

If the legs have lengths a and b, and the hypotenuse has length c, then

a2+b2=c2.
An image shows a right angle triangle ABC, which represents angle C is a 90 degree angle. AC represents as a base labeled, b, Leg, BC denotes as a height labeled, a Leg, and AB is labeled, c, hypotenuse.

Example 8 Using the Pythagorean Theorem: Screen Math

A bar graph shows the rising percentage of large screen TVs on store shelves.

Source: GAP Intelligence

Did you know that the size of a television screen refers to the length of its diagonal? If the length of the HDTV screen in Figure P.19 is 28 inches and its width is 15.7 inches, what is the size of the screen to the nearest inch?

Figure P.19

A photo shows a television with a screen size of length 28 inches and a width of 15.7 inches.

Solution

Figure P.19 shows that the length, width, and diagonal of the screen form a right triangle. The diagonal is the hypotenuse of the triangle. We use the Pythagorean Theorem with a=28, b=15.7, and solve for the screen size, c.

a2+b2=c2This is the Pythagorean Theorem.282+15.72=c2Let a=28 and b=15.7.784+246.49=c2282=2828=784 and15.72=15.715.7=246.49.c2=1030.49Add: 784+246.49=1030.49. We also reversedthe two sides.
c=1030.49 or  c=1030.49Apply the square root property.c32 or  c32Use a calculator and round to thenearest inch. 1030.49    = or   1030.49 ENTER

Because c represents the size of the screen, this dimension must be positive. We reject 32. Thus, the screen size of the HDTV is 32 inches.

Check Point 8

  • Figure P.20 shows the dimensions of an old TV screen. What is the size of the screen?

    Figure P.20

    A photo shows a television with a screen size of length 25.6 inches and a width of 19.2 inches.

Example 9 Dividing the Cost of a Yacht

A group of friends agrees to share the cost of a $50,000 yacht equally. Before the purchase is made, one more person joins the group. As a result, each person’s share is reduced by $2500. How many people were in the original group?

Solution

  1. Step 1 LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the number of people in the original group.
  2. Step 2 REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x.  Because one more person joined the original group, let

    x+1=the number of people in the final group.
  3. Step 3 WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. As a result of one more person joining the original group, each person’s share is reduced by $2500.

    The image shows a mathematical expression, start fraction 50,000 over x end fraction minus 2500 = start fraction 50,000 over x + 1 end fraction.
  4. Step 4 SOLVE THE EQUATION AND ANSWER THE QUESTION.

    50,000x2500=50,000x+1This is the equation that models theproblems conditions.x(x+1)(50,000x2500)=x(x+1)50,000x+1Multiply both sides by x(x+1),the LCD. x (x+1)50,000 x x(x+1)2500=x (x+1) 50,000 (x+1) Use the distributive property anddivide out common factors.50,000(x+1)2500x(x+1)=50,000xSimplify.50,000x+50,0002500x22500x=50,000xUse the distributive property.2500x2+47,500x+50,000=50,000xCombine like terms.2500x22500x+50,000=0Write the quadratic equation instandard form.2500(x2+x20)=0Factor out 2500.2500(x+5)(x4)=0Factor completely.x+5=0or x4=0Set each variable factor equal to zero.x=5x=4Solve the resulting equations.

    Because x represents the number of people in the original group, x cannot be negative. Thus, there were four people in the original group.

  5. Step 5 CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM.

     original cost per person=$50,0004=$12,500 final cost per person=$50,0005=$10,000

    We see that each person’s share is reduced by $12,500$10,000, or $2500, as specified by the problem’s conditions.

Check Point 9

  • A group of people share equally in a $5,000,000 lottery. Before the money is divided, three more winning ticket holders are declared. As a result, each person’s share is reduced by $375,000. How many people were in the original group of winners?

Objective 1: Use equations to solve problems

Problem Solving with Equations

  1. Objective 1Use equations to solve problems.

Watch Video

We have seen that a model is a mathematical representation of a real-world situation. In this section, we will be solving problems that are presented in English. This means that we must obtain models by translating from the ordinary language of English into the language of algebraic equations. To translate, however, we must understand the English prose and be familiar with the forms of algebraic language. Following are some general steps we will use in solving word problems:

Strategy for Solving Word Problems

  1. Step 1 Read the problem carefully several times until you can state in your own words what is given and what the problem is looking for. Let x (or any variable) represent one of the unknown quantities in the problem.

  2. Step 2 If necessary, write expressions for any other unknown quantities in the problem in terms of x.

  3. Step 3 Write an equation in x that models the verbal conditions of the problem.

  4. Step 4 Solve the equation and answer the problem’s question.

  5. Step 5 Check the solution in the original wording of the problem, not in the equation obtained from the words.

Example 1 Education Pays Off

The graph in Figure P.14 shows median yearly earnings in the United States by highest educational attainment. It certainly looks like you can make more money with more education. However, workers who make more money pay a higher percentage of income in taxes, so it is wise to consider after-tax earnings when deciding if that higher degree will pay off.

Figure P.14

A bar chart shows median earnings of full-time workers in the United States, by highest educational attainment.

Source: Education Pays 2019

The median yearly after-tax income of a full-time worker with a bachelor’s degree exceeds that of a full-time worker with an associate’s degree by $11 thousand. The median yearly after-tax income of a full-time worker with a master’s degree exceeds that of a full-time worker with an associate’s degree by $21 thousand. Combined, three full-time workers with each of these degrees earn $149 thousand after taxes. Find the median yearly after-tax income of full-time workers with each of these levels of education.

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We know something about after-tax incomes of full-time workers with bachelor’s degrees and master’s degrees: They exceed the after-tax income of a full-time worker with an associate’s degree by $11 thousand and $21 thousand, respectively. We will let

    x=the median yearly after-tax income of a full-time worker with an associates degree.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. Because a full-time worker with a bachelor’s degree earns $11 thousand more after taxes than a full-time worker with an associate’s degree, let

    x+11=the median yearly after-tax income of a full-time worker with a bachelors degree.

    Because a full-time worker with a master’s degree earns $21 thousand more after taxes than a full-time worker with an associate’s degree, let

    x+21=the median yearly after-tax income of a full-time worker with a master's degree.
  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. Combined, three full-time workers with each of these degrees earn $149 thousand.

    An equation with variable x to model the conditions.

  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    x+(x+11)+(x+21)=149This is the equation that models the problems conditions.3x+32=149Remove parentheses, regroup, and combine like terms.3x=117Subtract 32 from both sides.x=39Divide both sides by 3.

    Because we isolated the variable in the model and obtained x=39,

    median after-tax income with an associates degree=x=39median after-tax income with a bachelors degree=x+11=39+11=50median after-tax income with a masters degree=x+21=39+21=60.

    Full-time workers with associate’s degrees earn $39 thousand per year after taxes, full-time workers with bachelor’s degrees earn $50 thousand per year after taxes, and full-time workers with master’s degrees earn $60 thousand per year after taxes.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that combined, three full-time workers with each of these educational attainments earn $149 thousand after taxes. Using the incomes we determined in step 4, the sum is

    $39 thousand+$50 thousand+$60 thousand,or $149 thousand,

    which satisfies the problem’s conditions.

Check Point 1

  • The median yearly before-tax income of a full-time worker with a bachelor’s degree exceeds that of a full-time worker with an associate’s degree by $15 thousand. The median yearly before-tax income of a full-time worker with a master’s degree exceeds that of a full-time worker with an associate’s degree by $30 thousand. Combined, three full-time workers with each of these educational attainments earn $195 thousand before taxes. Find the median yearly before-tax income of full-time workers with each of these levels of education. (These incomes are illustrated by the bar graph on the previous page.)

Example 2 Modeling Attitudes of College Freshmen

Your author teaching math in 1969

Researchers have surveyed college freshmen every year since 1969. Figure P.15 shows that attitudes about some life goals have changed dramatically over the years. In particular, the freshman class of 2018 was more interested in making money than the freshmen of 1969 had been. In 1969, 42% of first-year college students considered “being well-off financially” essential or very important. For the period from 1969 through 2018, this percentage increased by approximately 0.8 each year. If this trend continues, by which year will all college freshmen consider “being well-off financially” essential or very important?

Figure P.15

A bar chart shows the life objectives of college freshmen, for the period from 19 69 to 20 18.

Source: Higher Education Research Institute

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We are interested in the year when all college freshmen, or 100% of the freshmen, will consider this life objective essential or very important. Let

    x = the number of years after 1969 when all freshmen will considerbeing    well-off financially essential or very important.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. There are no other unknown quantities to find, so we can skip this step.

  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS.

    The image shows a mathematical expression 42 + 0.8 x = 100
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    42+0.8x=100This is the equation that models the problems conditions.4242+0.8x=10042Subtract 42 from both sides.0.8x=58Simplify.0.8x0.8=580.8Divide both sides by 0.8.x=72.573Simplify and round to the nearestwhole number.

    Using current trends, by approximately 73 years after 1969, or in 2042, all freshmen will consider “being well-off financially” essential or very important.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that all freshmen (100%, represented by 100 using the model) will consider the objective essential or very important. Does this approximately occur if we increase the 1969 percentage, 42, by 0.8 each year for 73 years, our proposed solution?

    42+0.8(73)=42+58.4=100.4100

    This verifies that using trends shown in Figure P.15, all first-year college students will consider the objective of being well-off financially essential or very important approximately 73 years after 1969, or in 2042.

Check Point 2

  • Figure P.15 on the previous page shows that the freshman class of 2018 was less interested in developing a philosophy of life than the freshmen of 1969 had been. In 1969, 85% of the freshmen considered this objective essential or very important. Since then, this percentage has decreased by approximately 0.8 each year. If this trend continues, by which year will only 25% of college freshmen consider “developing a meaningful philosophy of life” essential or very important?

Example 3 Modeling Options for a Toll

It’s time to decide how you’re going to pay the bridge toll so that you can get to the beach. The first option requires purchasing a transponder for $20; with the transponder, you pay a reduced toll of $3.25 each time you cross the bridge. Your second option is toll-by-plate; with this option, you pay the full toll of $4.25 each time you cross the bridge plus a $3 administrative fee for each crossing. Find the number of times you would need to cross the bridge for the costs of the two options to be the same.

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. Let

    x=the number of times you cross the bridge.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. There are no other unknown quantities, so we can skip this step.

  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. The cost with the transponder is the cost of the transponder, $20, plus the toll, $3.25, times the number of crossings, x. For toll-by-plate, you must pay both the toll, $4.25, and the administrative fee, $3, each time you cross the bridge. The cost for toll-by-plate is the sum of the toll and the administrative fee, $4.25+$3 or $7.25, times the number of crossings, x. We want to know when these two costs are equal.

    The image shows a mathematical expression 7.25x = 20 + 3.25 x
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    7.25x=20+3.25xThis is the equation that models theproblems conditions.4x=20Subtract 3.25x from both sides.4x=5Divide both sides by 4.

    Because x represents the number of times you cross the bridge, the total cost of toll-by-plate is the same as the total cost with the transponder for 5 crossings.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that the cost of toll-by-plate should be the same as the cost with the transponder. Let’s see if they are the same with 5 bridge crossings per month.

    Expression to check the proposed solution.

    If you cross the bridge five times, both options cost $36.25. Thus the proposed solution, 5 crossings, satisfies the problem’s conditions.

Check Point 3

  • You drive up to a toll plaza and find booths with attendants where you can pay the toll by cash or credit card. With this option, the toll is $5 each time you cross the bridge. The attendant gives you the option of buying a bar-coded decal for $25; with the decal, you get 25% off the normal toll of $5 for each crossing. Find the number of times you would need to cross the bridge for the costs of the two options to be the same.

Example 4 A Price Reduction on Wireless Headphones

Your favorite electronics retailer is having a terrific sale on noise cancelling wireless headphones. (You’ll finally be able to listen to music in peace!) After a 40% price reduction, you purchase headphones for $192. What was the price of the headphones before the reduction?

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the price of the headphones prior to the reduction.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. There are no other unknown quantities to find, so we can skip this step.

  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. The headphones’ original price minus the 40% reduction is the reduced price, $192.

    The image shows a mathematical expression x minus 0.4 x = 192.
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    The solution of an equation.

    The headphones’ price before the reduction was $320.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The price before the reduction, $320, minus the 40% reduction should equal the reduced price given in the original wording, $192:

    32040% of 320=3200.4(320)=320128=192.

    This verifies that the headphones’ price before the reduction was $320.

Check Point 4

  • After a 30% price reduction, you purchase a new laptop for $840. What was the laptop’s price before the reduction?

Our next example is about simple interest. Simple interest involves interest calculated only on the amount of money that we invest, called the principal. The formula I=Pr is used to find the simple interest, I, earned for one year when the principal, P, is invested at an annual interest rate, r. Dual investment problems involve different amounts of money in two or more investments, each paying a different rate.

Example 5 Solving a Dual Investment Problem

Your grandmother needs your help. She has $150,000 to invest. Part of this money is to be invested in a money market account paying 1.5% annual interest. The rest of this money is to be invested in a certificate of deposit paying 1.3% annual interest. She told you that she requires $2000 per year in extra income from the combination of these investments. How much money should be placed in each investment?

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the amount invested in the money market account at 1.5%.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. The other quantity that we seek is the amount invested at 1.3% in the certificate of deposit. Because the total amount Grandma has to invest is $150,000 and we already used up x,

    150,000x=the amount invested in the certificate of deposit at 1.3%.
  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. Because Grandma requires $2000 in total interest, the interest for the two investments combined must be $2000. Interest is Pr or rP for each investment.

    The image shows a mathematical expression 0.015 x + 0.013 left parenthesis 150,000 minus x right parenthesis = 20 00
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    A mathematical expression steps to solve an equation to find the value of x.

    Thus,

    the amount invested at1.5%=x=25,000the amount invested at 1.3%=150,00025,000=125,000

    Grandma should invest $25,000 at 1.5% and $125,000 at 1.3%.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that the total interest from the dual investments should be $2000. Can Grandma count on $2000 interest? The interest earned on $25,000 at 1.5% is ($25,000) (0.015), or $375. The interest earned on $125,000 at 1.3% is ($125,000)(0.013), or $1625. The total interest is $375+$1625, or $2000, exactly as it should be. You’ve made your grandmother happy. (Now if you would just visit her more often …)

Check Point 5

  • You inherited $50,000 with the stipulation that for the first year the money had to be invested in two accounts paying 0.9% and 1.1% annual interest. How much did you invest at each rate if the total interest earned for the year was $515?

Solving geometry problems usually requires a knowledge of basic geometric ideas and formulas. Formulas for area, perimeter, and volume are given in Table P.6.

Table P.6 Common Formulas for Area, Perimeter, and Volume

An illustration shows common formulas for area, perimeter, and volume of different shapes.
Table P.6 Full Alternative Text

We will be using the formula for the perimeter of a rectangle, P=2l+2w, in our next example. The formula states that a rectangle’s perimeter is the sum of twice its length and twice its width.

Example 6 Finding the Dimensions of an American Football Field

The length of an American football field is 200 feet more than the width. If the perimeter of the field is 1040 feet, what are its dimensions?

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We know something about the length; the length is 200 feet more than the width. We will let

    x=the width.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. Because the length is 200 feet more than the width, we add 200 to the width to represent the length. Thus,

    x+200=the length.

    Figure P.16 illustrates an American football field and its dimensions.

    Figure P.16 An American football field

    An illustration shows an American football field with its width as x and length as x + 20 0.
  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. Because the perimeter of the field is 1040 feet,

    The image shows the mathematical expression 2 left parenthesis x + 20 0 right parenthesis + 2 x = 1040.

  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    Steps to solve an equation.

    Thus,

    width=x=160.length=x+200=160+200=360.

    The dimensions of an American football field are 160 feet by 360 feet. (The 360-foot length is usually described as 120 yards.)

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The perimeter of the football field using the dimensions that we found is

    2(160 feet)+2(360 feet)=320 feet+720 feet=1040 feet.

    Because the problem’s wording tells us that the perimeter is 1040 feet, our dimensions are correct.

Check Point 6

  • The length of a rectangular basketball court is 44 feet more than the width. If the perimeter of the basketball court is 288 feet, what are its dimensions?

We will use the formula for the area of a rectangle, A=lw, in our next example. The formula states that a rectangle’s area is the product of its length and its width.

Example 7 Solving a Problem Involving Landscape Design

A rectangular garden measures 80 feet by 60 feet. A large path of uniform width is to be added along both shorter sides and one longer side of the garden. The landscape designer doing the work wants to double the garden’s area with the addition of this path. How wide should the path be?

Solution

  1. Step 1 LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the width of the path.

    The situation is illustrated in Figure P.17. The figure shows the original 80-by-60 foot rectangular garden and the path of width x added along both shorter sides and one longer side.

    Figure P.17 The garden’s area is to be doubled by adding the path.

    An image shows a rectangular garden is surrounded by a path that has a length of 80 + 2 x and a width of 60 + x.
  2. Step 2 REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. Because the path is added along both shorter sides and one longer side, Figure P.17 shows that

     80+2x=the length of the new, expanded rectangle 60+x=the width of the new, expanded rectangle.

  3. Step 3 WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. The area of the rectangle must be doubled by the addition of the path.

    Left parenthesis 80 + 2 x right parenthesis, left parenthesis 60 + x right parenthesis = 2 times 80 times 60.
  4. Step 4 SOLVE THE EQUATION AND ANSWER THE QUESTION.

    (80+2x)(60+x)=28060This is the equation that modelsthe problems conditions.4800+200x+2x2=9600Multiply. Use FOIL on the left side.2x2+200x4800=0Subtract 9600 from both sidesand write the quadratic equationin standard form.2(x2+100x2400)=0Factor out 2, the GCF.2(x20)(x+120)=0Factor the trinomial.x20=0or x+120=0Set each variable factor equal to 0.x=20x=120Solve for x.

    The path cannot have a negative width. Because 120 is geometrically impossible, we use x=20. The width of the path should be 20 feet.

  5. Step 5 CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. Has the landscape architect doubled the garden’s area with the 20-foot-wide path? The area of the garden is 80 feet times 60 feet, or 4800 square feet. Because 80+2x and 60+x represent the length and width of the expanded rectangle,

    80+2x=80+220 =120 feet is the expanded rectangle's length.60+2x=60+20=80 feet is the expanded rectangle's width.

    The area of the expanded rectangle is 120 feet times 80 feet, or 9600 square feet. This is double the area of the garden, 4800 square feet, as specified by the problem’s conditions.

Check Point 7

  • A rectangular garden measures 16 feet by 12 feet. A path of uniform width is to be added so as to surround the entire garden. The landscape artist doing the work wants the garden and path to cover an area of 320 square feet. How wide should the path be?

In our next example, we will be using the Pythagorean Theorem to obtain a mathematical model. The ancient Greek philosopher and mathematician Pythagoras (approximately 582–500 b.c.) founded a school whose motto was “All is number.” Pythagoras is best remembered for his work with the right triangle, a triangle with one angle measuring 90°. The side opposite the 90° angle is called the hypotenuse. The other sides are called legs. Pythagoras found that if he constructed squares on each of the legs, as well as a larger square on the hypotenuse, the sum of the areas of the smaller squares is equal to the area of the larger square. This is illustrated in Figure P.18.

Figure P.18The area of the large square equals the sum of the areas of the smaller squares.

An illustration shows a right triangle of base 3, height 4, and hypotenuse 5.
Figure P. 18 Full Alternative Text

This relationship is usually stated in terms of the lengths of the three sides of a right triangle and is called the Pythagorean Theorem.

The Pythagorean Theorem

The sum of the squares of the lengths of the legs of a right triangle equals the square of the length of the hypotenuse.

If the legs have lengths a and b, and the hypotenuse has length c, then

a2+b2=c2.
An image shows a right angle triangle ABC, which represents angle C is a 90 degree angle. AC represents as a base labeled, b, Leg, BC denotes as a height labeled, a Leg, and AB is labeled, c, hypotenuse.

Example 8 Using the Pythagorean Theorem: Screen Math

A bar graph shows the rising percentage of large screen TVs on store shelves.

Source: GAP Intelligence

Did you know that the size of a television screen refers to the length of its diagonal? If the length of the HDTV screen in Figure P.19 is 28 inches and its width is 15.7 inches, what is the size of the screen to the nearest inch?

Figure P.19

A photo shows a television with a screen size of length 28 inches and a width of 15.7 inches.

Solution

Figure P.19 shows that the length, width, and diagonal of the screen form a right triangle. The diagonal is the hypotenuse of the triangle. We use the Pythagorean Theorem with a=28, b=15.7, and solve for the screen size, c.

a2+b2=c2This is the Pythagorean Theorem.282+15.72=c2Let a=28 and b=15.7.784+246.49=c2282=2828=784 and15.72=15.715.7=246.49.c2=1030.49Add: 784+246.49=1030.49. We also reversedthe two sides.
c=1030.49 or  c=1030.49Apply the square root property.c32 or  c32Use a calculator and round to thenearest inch. 1030.49    = or   1030.49 ENTER

Because c represents the size of the screen, this dimension must be positive. We reject 32. Thus, the screen size of the HDTV is 32 inches.

Check Point 8

  • Figure P.20 shows the dimensions of an old TV screen. What is the size of the screen?

    Figure P.20

    A photo shows a television with a screen size of length 25.6 inches and a width of 19.2 inches.

Example 9 Dividing the Cost of a Yacht

A group of friends agrees to share the cost of a $50,000 yacht equally. Before the purchase is made, one more person joins the group. As a result, each person’s share is reduced by $2500. How many people were in the original group?

Solution

  1. Step 1 LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the number of people in the original group.
  2. Step 2 REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x.  Because one more person joined the original group, let

    x+1=the number of people in the final group.
  3. Step 3 WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. As a result of one more person joining the original group, each person’s share is reduced by $2500.

    The image shows a mathematical expression, start fraction 50,000 over x end fraction minus 2500 = start fraction 50,000 over x + 1 end fraction.
  4. Step 4 SOLVE THE EQUATION AND ANSWER THE QUESTION.

    50,000x2500=50,000x+1This is the equation that models theproblems conditions.x(x+1)(50,000x2500)=x(x+1)50,000x+1Multiply both sides by x(x+1),the LCD. x (x+1)50,000 x x(x+1)2500=x (x+1) 50,000 (x+1) Use the distributive property anddivide out common factors.50,000(x+1)2500x(x+1)=50,000xSimplify.50,000x+50,0002500x22500x=50,000xUse the distributive property.2500x2+47,500x+50,000=50,000xCombine like terms.2500x22500x+50,000=0Write the quadratic equation instandard form.2500(x2+x20)=0Factor out 2500.2500(x+5)(x4)=0Factor completely.x+5=0or x4=0Set each variable factor equal to zero.x=5x=4Solve the resulting equations.

    Because x represents the number of people in the original group, x cannot be negative. Thus, there were four people in the original group.

  5. Step 5 CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM.

     original cost per person=$50,0004=$12,500 final cost per person=$50,0005=$10,000

    We see that each person’s share is reduced by $12,500$10,000, or $2500, as specified by the problem’s conditions.

Check Point 9

  • A group of people share equally in a $5,000,000 lottery. Before the money is divided, three more winning ticket holders are declared. As a result, each person’s share is reduced by $375,000. How many people were in the original group of winners?

Objective 1: Use equations to solve problems

Problem Solving with Equations

  1. Objective 1Use equations to solve problems.

Watch Video

We have seen that a model is a mathematical representation of a real-world situation. In this section, we will be solving problems that are presented in English. This means that we must obtain models by translating from the ordinary language of English into the language of algebraic equations. To translate, however, we must understand the English prose and be familiar with the forms of algebraic language. Following are some general steps we will use in solving word problems:

Strategy for Solving Word Problems

  1. Step 1 Read the problem carefully several times until you can state in your own words what is given and what the problem is looking for. Let x (or any variable) represent one of the unknown quantities in the problem.

  2. Step 2 If necessary, write expressions for any other unknown quantities in the problem in terms of x.

  3. Step 3 Write an equation in x that models the verbal conditions of the problem.

  4. Step 4 Solve the equation and answer the problem’s question.

  5. Step 5 Check the solution in the original wording of the problem, not in the equation obtained from the words.

Example 1 Education Pays Off

The graph in Figure P.14 shows median yearly earnings in the United States by highest educational attainment. It certainly looks like you can make more money with more education. However, workers who make more money pay a higher percentage of income in taxes, so it is wise to consider after-tax earnings when deciding if that higher degree will pay off.

Figure P.14

A bar chart shows median earnings of full-time workers in the United States, by highest educational attainment.

Source: Education Pays 2019

The median yearly after-tax income of a full-time worker with a bachelor’s degree exceeds that of a full-time worker with an associate’s degree by $11 thousand. The median yearly after-tax income of a full-time worker with a master’s degree exceeds that of a full-time worker with an associate’s degree by $21 thousand. Combined, three full-time workers with each of these degrees earn $149 thousand after taxes. Find the median yearly after-tax income of full-time workers with each of these levels of education.

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We know something about after-tax incomes of full-time workers with bachelor’s degrees and master’s degrees: They exceed the after-tax income of a full-time worker with an associate’s degree by $11 thousand and $21 thousand, respectively. We will let

    x=the median yearly after-tax income of a full-time worker with an associates degree.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. Because a full-time worker with a bachelor’s degree earns $11 thousand more after taxes than a full-time worker with an associate’s degree, let

    x+11=the median yearly after-tax income of a full-time worker with a bachelors degree.

    Because a full-time worker with a master’s degree earns $21 thousand more after taxes than a full-time worker with an associate’s degree, let

    x+21=the median yearly after-tax income of a full-time worker with a master's degree.
  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. Combined, three full-time workers with each of these degrees earn $149 thousand.

    An equation with variable x to model the conditions.

  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    x+(x+11)+(x+21)=149This is the equation that models the problems conditions.3x+32=149Remove parentheses, regroup, and combine like terms.3x=117Subtract 32 from both sides.x=39Divide both sides by 3.

    Because we isolated the variable in the model and obtained x=39,

    median after-tax income with an associates degree=x=39median after-tax income with a bachelors degree=x+11=39+11=50median after-tax income with a masters degree=x+21=39+21=60.

    Full-time workers with associate’s degrees earn $39 thousand per year after taxes, full-time workers with bachelor’s degrees earn $50 thousand per year after taxes, and full-time workers with master’s degrees earn $60 thousand per year after taxes.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that combined, three full-time workers with each of these educational attainments earn $149 thousand after taxes. Using the incomes we determined in step 4, the sum is

    $39 thousand+$50 thousand+$60 thousand,or $149 thousand,

    which satisfies the problem’s conditions.

Check Point 1

  • The median yearly before-tax income of a full-time worker with a bachelor’s degree exceeds that of a full-time worker with an associate’s degree by $15 thousand. The median yearly before-tax income of a full-time worker with a master’s degree exceeds that of a full-time worker with an associate’s degree by $30 thousand. Combined, three full-time workers with each of these educational attainments earn $195 thousand before taxes. Find the median yearly before-tax income of full-time workers with each of these levels of education. (These incomes are illustrated by the bar graph on the previous page.)

Example 2 Modeling Attitudes of College Freshmen

Your author teaching math in 1969

Researchers have surveyed college freshmen every year since 1969. Figure P.15 shows that attitudes about some life goals have changed dramatically over the years. In particular, the freshman class of 2018 was more interested in making money than the freshmen of 1969 had been. In 1969, 42% of first-year college students considered “being well-off financially” essential or very important. For the period from 1969 through 2018, this percentage increased by approximately 0.8 each year. If this trend continues, by which year will all college freshmen consider “being well-off financially” essential or very important?

Figure P.15

A bar chart shows the life objectives of college freshmen, for the period from 19 69 to 20 18.

Source: Higher Education Research Institute

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We are interested in the year when all college freshmen, or 100% of the freshmen, will consider this life objective essential or very important. Let

    x = the number of years after 1969 when all freshmen will considerbeing    well-off financially essential or very important.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. There are no other unknown quantities to find, so we can skip this step.

  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS.

    The image shows a mathematical expression 42 + 0.8 x = 100
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    42+0.8x=100This is the equation that models the problems conditions.4242+0.8x=10042Subtract 42 from both sides.0.8x=58Simplify.0.8x0.8=580.8Divide both sides by 0.8.x=72.573Simplify and round to the nearestwhole number.

    Using current trends, by approximately 73 years after 1969, or in 2042, all freshmen will consider “being well-off financially” essential or very important.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that all freshmen (100%, represented by 100 using the model) will consider the objective essential or very important. Does this approximately occur if we increase the 1969 percentage, 42, by 0.8 each year for 73 years, our proposed solution?

    42+0.8(73)=42+58.4=100.4100

    This verifies that using trends shown in Figure P.15, all first-year college students will consider the objective of being well-off financially essential or very important approximately 73 years after 1969, or in 2042.

Check Point 2

  • Figure P.15 on the previous page shows that the freshman class of 2018 was less interested in developing a philosophy of life than the freshmen of 1969 had been. In 1969, 85% of the freshmen considered this objective essential or very important. Since then, this percentage has decreased by approximately 0.8 each year. If this trend continues, by which year will only 25% of college freshmen consider “developing a meaningful philosophy of life” essential or very important?

Example 3 Modeling Options for a Toll

It’s time to decide how you’re going to pay the bridge toll so that you can get to the beach. The first option requires purchasing a transponder for $20; with the transponder, you pay a reduced toll of $3.25 each time you cross the bridge. Your second option is toll-by-plate; with this option, you pay the full toll of $4.25 each time you cross the bridge plus a $3 administrative fee for each crossing. Find the number of times you would need to cross the bridge for the costs of the two options to be the same.

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. Let

    x=the number of times you cross the bridge.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. There are no other unknown quantities, so we can skip this step.

  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. The cost with the transponder is the cost of the transponder, $20, plus the toll, $3.25, times the number of crossings, x. For toll-by-plate, you must pay both the toll, $4.25, and the administrative fee, $3, each time you cross the bridge. The cost for toll-by-plate is the sum of the toll and the administrative fee, $4.25+$3 or $7.25, times the number of crossings, x. We want to know when these two costs are equal.

    The image shows a mathematical expression 7.25x = 20 + 3.25 x
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    7.25x=20+3.25xThis is the equation that models theproblems conditions.4x=20Subtract 3.25x from both sides.4x=5Divide both sides by 4.

    Because x represents the number of times you cross the bridge, the total cost of toll-by-plate is the same as the total cost with the transponder for 5 crossings.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that the cost of toll-by-plate should be the same as the cost with the transponder. Let’s see if they are the same with 5 bridge crossings per month.

    Expression to check the proposed solution.

    If you cross the bridge five times, both options cost $36.25. Thus the proposed solution, 5 crossings, satisfies the problem’s conditions.

Check Point 3

  • You drive up to a toll plaza and find booths with attendants where you can pay the toll by cash or credit card. With this option, the toll is $5 each time you cross the bridge. The attendant gives you the option of buying a bar-coded decal for $25; with the decal, you get 25% off the normal toll of $5 for each crossing. Find the number of times you would need to cross the bridge for the costs of the two options to be the same.

Example 4 A Price Reduction on Wireless Headphones

Your favorite electronics retailer is having a terrific sale on noise cancelling wireless headphones. (You’ll finally be able to listen to music in peace!) After a 40% price reduction, you purchase headphones for $192. What was the price of the headphones before the reduction?

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the price of the headphones prior to the reduction.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. There are no other unknown quantities to find, so we can skip this step.

  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. The headphones’ original price minus the 40% reduction is the reduced price, $192.

    The image shows a mathematical expression x minus 0.4 x = 192.
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    The solution of an equation.

    The headphones’ price before the reduction was $320.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The price before the reduction, $320, minus the 40% reduction should equal the reduced price given in the original wording, $192:

    32040% of 320=3200.4(320)=320128=192.

    This verifies that the headphones’ price before the reduction was $320.

Check Point 4

  • After a 30% price reduction, you purchase a new laptop for $840. What was the laptop’s price before the reduction?

Our next example is about simple interest. Simple interest involves interest calculated only on the amount of money that we invest, called the principal. The formula I=Pr is used to find the simple interest, I, earned for one year when the principal, P, is invested at an annual interest rate, r. Dual investment problems involve different amounts of money in two or more investments, each paying a different rate.

Example 5 Solving a Dual Investment Problem

Your grandmother needs your help. She has $150,000 to invest. Part of this money is to be invested in a money market account paying 1.5% annual interest. The rest of this money is to be invested in a certificate of deposit paying 1.3% annual interest. She told you that she requires $2000 per year in extra income from the combination of these investments. How much money should be placed in each investment?

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the amount invested in the money market account at 1.5%.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. The other quantity that we seek is the amount invested at 1.3% in the certificate of deposit. Because the total amount Grandma has to invest is $150,000 and we already used up x,

    150,000x=the amount invested in the certificate of deposit at 1.3%.
  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. Because Grandma requires $2000 in total interest, the interest for the two investments combined must be $2000. Interest is Pr or rP for each investment.

    The image shows a mathematical expression 0.015 x + 0.013 left parenthesis 150,000 minus x right parenthesis = 20 00
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    A mathematical expression steps to solve an equation to find the value of x.

    Thus,

    the amount invested at1.5%=x=25,000the amount invested at 1.3%=150,00025,000=125,000

    Grandma should invest $25,000 at 1.5% and $125,000 at 1.3%.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that the total interest from the dual investments should be $2000. Can Grandma count on $2000 interest? The interest earned on $25,000 at 1.5% is ($25,000) (0.015), or $375. The interest earned on $125,000 at 1.3% is ($125,000)(0.013), or $1625. The total interest is $375+$1625, or $2000, exactly as it should be. You’ve made your grandmother happy. (Now if you would just visit her more often …)

Check Point 5

  • You inherited $50,000 with the stipulation that for the first year the money had to be invested in two accounts paying 0.9% and 1.1% annual interest. How much did you invest at each rate if the total interest earned for the year was $515?

Solving geometry problems usually requires a knowledge of basic geometric ideas and formulas. Formulas for area, perimeter, and volume are given in Table P.6.

Table P.6 Common Formulas for Area, Perimeter, and Volume

An illustration shows common formulas for area, perimeter, and volume of different shapes.
Table P.6 Full Alternative Text

We will be using the formula for the perimeter of a rectangle, P=2l+2w, in our next example. The formula states that a rectangle’s perimeter is the sum of twice its length and twice its width.

Example 6 Finding the Dimensions of an American Football Field

The length of an American football field is 200 feet more than the width. If the perimeter of the field is 1040 feet, what are its dimensions?

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We know something about the length; the length is 200 feet more than the width. We will let

    x=the width.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. Because the length is 200 feet more than the width, we add 200 to the width to represent the length. Thus,

    x+200=the length.

    Figure P.16 illustrates an American football field and its dimensions.

    Figure P.16 An American football field

    An illustration shows an American football field with its width as x and length as x + 20 0.
  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. Because the perimeter of the field is 1040 feet,

    The image shows the mathematical expression 2 left parenthesis x + 20 0 right parenthesis + 2 x = 1040.

  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    Steps to solve an equation.

    Thus,

    width=x=160.length=x+200=160+200=360.

    The dimensions of an American football field are 160 feet by 360 feet. (The 360-foot length is usually described as 120 yards.)

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The perimeter of the football field using the dimensions that we found is

    2(160 feet)+2(360 feet)=320 feet+720 feet=1040 feet.

    Because the problem’s wording tells us that the perimeter is 1040 feet, our dimensions are correct.

Check Point 6

  • The length of a rectangular basketball court is 44 feet more than the width. If the perimeter of the basketball court is 288 feet, what are its dimensions?

We will use the formula for the area of a rectangle, A=lw, in our next example. The formula states that a rectangle’s area is the product of its length and its width.

Example 7 Solving a Problem Involving Landscape Design

A rectangular garden measures 80 feet by 60 feet. A large path of uniform width is to be added along both shorter sides and one longer side of the garden. The landscape designer doing the work wants to double the garden’s area with the addition of this path. How wide should the path be?

Solution

  1. Step 1 LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the width of the path.

    The situation is illustrated in Figure P.17. The figure shows the original 80-by-60 foot rectangular garden and the path of width x added along both shorter sides and one longer side.

    Figure P.17 The garden’s area is to be doubled by adding the path.

    An image shows a rectangular garden is surrounded by a path that has a length of 80 + 2 x and a width of 60 + x.
  2. Step 2 REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. Because the path is added along both shorter sides and one longer side, Figure P.17 shows that

     80+2x=the length of the new, expanded rectangle 60+x=the width of the new, expanded rectangle.

  3. Step 3 WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. The area of the rectangle must be doubled by the addition of the path.

    Left parenthesis 80 + 2 x right parenthesis, left parenthesis 60 + x right parenthesis = 2 times 80 times 60.
  4. Step 4 SOLVE THE EQUATION AND ANSWER THE QUESTION.

    (80+2x)(60+x)=28060This is the equation that modelsthe problems conditions.4800+200x+2x2=9600Multiply. Use FOIL on the left side.2x2+200x4800=0Subtract 9600 from both sidesand write the quadratic equationin standard form.2(x2+100x2400)=0Factor out 2, the GCF.2(x20)(x+120)=0Factor the trinomial.x20=0or x+120=0Set each variable factor equal to 0.x=20x=120Solve for x.

    The path cannot have a negative width. Because 120 is geometrically impossible, we use x=20. The width of the path should be 20 feet.

  5. Step 5 CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. Has the landscape architect doubled the garden’s area with the 20-foot-wide path? The area of the garden is 80 feet times 60 feet, or 4800 square feet. Because 80+2x and 60+x represent the length and width of the expanded rectangle,

    80+2x=80+220 =120 feet is the expanded rectangle's length.60+2x=60+20=80 feet is the expanded rectangle's width.

    The area of the expanded rectangle is 120 feet times 80 feet, or 9600 square feet. This is double the area of the garden, 4800 square feet, as specified by the problem’s conditions.

Check Point 7

  • A rectangular garden measures 16 feet by 12 feet. A path of uniform width is to be added so as to surround the entire garden. The landscape artist doing the work wants the garden and path to cover an area of 320 square feet. How wide should the path be?

In our next example, we will be using the Pythagorean Theorem to obtain a mathematical model. The ancient Greek philosopher and mathematician Pythagoras (approximately 582–500 b.c.) founded a school whose motto was “All is number.” Pythagoras is best remembered for his work with the right triangle, a triangle with one angle measuring 90°. The side opposite the 90° angle is called the hypotenuse. The other sides are called legs. Pythagoras found that if he constructed squares on each of the legs, as well as a larger square on the hypotenuse, the sum of the areas of the smaller squares is equal to the area of the larger square. This is illustrated in Figure P.18.

Figure P.18The area of the large square equals the sum of the areas of the smaller squares.

An illustration shows a right triangle of base 3, height 4, and hypotenuse 5.
Figure P. 18 Full Alternative Text

This relationship is usually stated in terms of the lengths of the three sides of a right triangle and is called the Pythagorean Theorem.

The Pythagorean Theorem

The sum of the squares of the lengths of the legs of a right triangle equals the square of the length of the hypotenuse.

If the legs have lengths a and b, and the hypotenuse has length c, then

a2+b2=c2.
An image shows a right angle triangle ABC, which represents angle C is a 90 degree angle. AC represents as a base labeled, b, Leg, BC denotes as a height labeled, a Leg, and AB is labeled, c, hypotenuse.

Example 8 Using the Pythagorean Theorem: Screen Math

A bar graph shows the rising percentage of large screen TVs on store shelves.

Source: GAP Intelligence

Did you know that the size of a television screen refers to the length of its diagonal? If the length of the HDTV screen in Figure P.19 is 28 inches and its width is 15.7 inches, what is the size of the screen to the nearest inch?

Figure P.19

A photo shows a television with a screen size of length 28 inches and a width of 15.7 inches.

Solution

Figure P.19 shows that the length, width, and diagonal of the screen form a right triangle. The diagonal is the hypotenuse of the triangle. We use the Pythagorean Theorem with a=28, b=15.7, and solve for the screen size, c.

a2+b2=c2This is the Pythagorean Theorem.282+15.72=c2Let a=28 and b=15.7.784+246.49=c2282=2828=784 and15.72=15.715.7=246.49.c2=1030.49Add: 784+246.49=1030.49. We also reversedthe two sides.
c=1030.49 or  c=1030.49Apply the square root property.c32 or  c32Use a calculator and round to thenearest inch. 1030.49    = or   1030.49 ENTER

Because c represents the size of the screen, this dimension must be positive. We reject 32. Thus, the screen size of the HDTV is 32 inches.

Check Point 8

  • Figure P.20 shows the dimensions of an old TV screen. What is the size of the screen?

    Figure P.20

    A photo shows a television with a screen size of length 25.6 inches and a width of 19.2 inches.

Example 9 Dividing the Cost of a Yacht

A group of friends agrees to share the cost of a $50,000 yacht equally. Before the purchase is made, one more person joins the group. As a result, each person’s share is reduced by $2500. How many people were in the original group?

Solution

  1. Step 1 LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the number of people in the original group.
  2. Step 2 REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x.  Because one more person joined the original group, let

    x+1=the number of people in the final group.
  3. Step 3 WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. As a result of one more person joining the original group, each person’s share is reduced by $2500.

    The image shows a mathematical expression, start fraction 50,000 over x end fraction minus 2500 = start fraction 50,000 over x + 1 end fraction.
  4. Step 4 SOLVE THE EQUATION AND ANSWER THE QUESTION.

    50,000x2500=50,000x+1This is the equation that models theproblems conditions.x(x+1)(50,000x2500)=x(x+1)50,000x+1Multiply both sides by x(x+1),the LCD. x (x+1)50,000 x x(x+1)2500=x (x+1) 50,000 (x+1) Use the distributive property anddivide out common factors.50,000(x+1)2500x(x+1)=50,000xSimplify.50,000x+50,0002500x22500x=50,000xUse the distributive property.2500x2+47,500x+50,000=50,000xCombine like terms.2500x22500x+50,000=0Write the quadratic equation instandard form.2500(x2+x20)=0Factor out 2500.2500(x+5)(x4)=0Factor completely.x+5=0or x4=0Set each variable factor equal to zero.x=5x=4Solve the resulting equations.

    Because x represents the number of people in the original group, x cannot be negative. Thus, there were four people in the original group.

  5. Step 5 CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM.

     original cost per person=$50,0004=$12,500 final cost per person=$50,0005=$10,000

    We see that each person’s share is reduced by $12,500$10,000, or $2500, as specified by the problem’s conditions.

Check Point 9

  • A group of people share equally in a $5,000,000 lottery. Before the money is divided, three more winning ticket holders are declared. As a result, each person’s share is reduced by $375,000. How many people were in the original group of winners?

Objective 1: Use equations to solve problems

Problem Solving with Equations

  1. Objective 1Use equations to solve problems.

Watch Video

We have seen that a model is a mathematical representation of a real-world situation. In this section, we will be solving problems that are presented in English. This means that we must obtain models by translating from the ordinary language of English into the language of algebraic equations. To translate, however, we must understand the English prose and be familiar with the forms of algebraic language. Following are some general steps we will use in solving word problems:

Strategy for Solving Word Problems

  1. Step 1 Read the problem carefully several times until you can state in your own words what is given and what the problem is looking for. Let x (or any variable) represent one of the unknown quantities in the problem.

  2. Step 2 If necessary, write expressions for any other unknown quantities in the problem in terms of x.

  3. Step 3 Write an equation in x that models the verbal conditions of the problem.

  4. Step 4 Solve the equation and answer the problem’s question.

  5. Step 5 Check the solution in the original wording of the problem, not in the equation obtained from the words.

Example 1 Education Pays Off

The graph in Figure P.14 shows median yearly earnings in the United States by highest educational attainment. It certainly looks like you can make more money with more education. However, workers who make more money pay a higher percentage of income in taxes, so it is wise to consider after-tax earnings when deciding if that higher degree will pay off.

Figure P.14

A bar chart shows median earnings of full-time workers in the United States, by highest educational attainment.

Source: Education Pays 2019

The median yearly after-tax income of a full-time worker with a bachelor’s degree exceeds that of a full-time worker with an associate’s degree by $11 thousand. The median yearly after-tax income of a full-time worker with a master’s degree exceeds that of a full-time worker with an associate’s degree by $21 thousand. Combined, three full-time workers with each of these degrees earn $149 thousand after taxes. Find the median yearly after-tax income of full-time workers with each of these levels of education.

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We know something about after-tax incomes of full-time workers with bachelor’s degrees and master’s degrees: They exceed the after-tax income of a full-time worker with an associate’s degree by $11 thousand and $21 thousand, respectively. We will let

    x=the median yearly after-tax income of a full-time worker with an associates degree.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. Because a full-time worker with a bachelor’s degree earns $11 thousand more after taxes than a full-time worker with an associate’s degree, let

    x+11=the median yearly after-tax income of a full-time worker with a bachelors degree.

    Because a full-time worker with a master’s degree earns $21 thousand more after taxes than a full-time worker with an associate’s degree, let

    x+21=the median yearly after-tax income of a full-time worker with a master's degree.
  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. Combined, three full-time workers with each of these degrees earn $149 thousand.

    An equation with variable x to model the conditions.

  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    x+(x+11)+(x+21)=149This is the equation that models the problems conditions.3x+32=149Remove parentheses, regroup, and combine like terms.3x=117Subtract 32 from both sides.x=39Divide both sides by 3.

    Because we isolated the variable in the model and obtained x=39,

    median after-tax income with an associates degree=x=39median after-tax income with a bachelors degree=x+11=39+11=50median after-tax income with a masters degree=x+21=39+21=60.

    Full-time workers with associate’s degrees earn $39 thousand per year after taxes, full-time workers with bachelor’s degrees earn $50 thousand per year after taxes, and full-time workers with master’s degrees earn $60 thousand per year after taxes.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that combined, three full-time workers with each of these educational attainments earn $149 thousand after taxes. Using the incomes we determined in step 4, the sum is

    $39 thousand+$50 thousand+$60 thousand,or $149 thousand,

    which satisfies the problem’s conditions.

Check Point 1

  • The median yearly before-tax income of a full-time worker with a bachelor’s degree exceeds that of a full-time worker with an associate’s degree by $15 thousand. The median yearly before-tax income of a full-time worker with a master’s degree exceeds that of a full-time worker with an associate’s degree by $30 thousand. Combined, three full-time workers with each of these educational attainments earn $195 thousand before taxes. Find the median yearly before-tax income of full-time workers with each of these levels of education. (These incomes are illustrated by the bar graph on the previous page.)

Example 2 Modeling Attitudes of College Freshmen

Your author teaching math in 1969

Researchers have surveyed college freshmen every year since 1969. Figure P.15 shows that attitudes about some life goals have changed dramatically over the years. In particular, the freshman class of 2018 was more interested in making money than the freshmen of 1969 had been. In 1969, 42% of first-year college students considered “being well-off financially” essential or very important. For the period from 1969 through 2018, this percentage increased by approximately 0.8 each year. If this trend continues, by which year will all college freshmen consider “being well-off financially” essential or very important?

Figure P.15

A bar chart shows the life objectives of college freshmen, for the period from 19 69 to 20 18.

Source: Higher Education Research Institute

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We are interested in the year when all college freshmen, or 100% of the freshmen, will consider this life objective essential or very important. Let

    x = the number of years after 1969 when all freshmen will considerbeing    well-off financially essential or very important.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. There are no other unknown quantities to find, so we can skip this step.

  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS.

    The image shows a mathematical expression 42 + 0.8 x = 100
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    42+0.8x=100This is the equation that models the problems conditions.4242+0.8x=10042Subtract 42 from both sides.0.8x=58Simplify.0.8x0.8=580.8Divide both sides by 0.8.x=72.573Simplify and round to the nearestwhole number.

    Using current trends, by approximately 73 years after 1969, or in 2042, all freshmen will consider “being well-off financially” essential or very important.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that all freshmen (100%, represented by 100 using the model) will consider the objective essential or very important. Does this approximately occur if we increase the 1969 percentage, 42, by 0.8 each year for 73 years, our proposed solution?

    42+0.8(73)=42+58.4=100.4100

    This verifies that using trends shown in Figure P.15, all first-year college students will consider the objective of being well-off financially essential or very important approximately 73 years after 1969, or in 2042.

Check Point 2

  • Figure P.15 on the previous page shows that the freshman class of 2018 was less interested in developing a philosophy of life than the freshmen of 1969 had been. In 1969, 85% of the freshmen considered this objective essential or very important. Since then, this percentage has decreased by approximately 0.8 each year. If this trend continues, by which year will only 25% of college freshmen consider “developing a meaningful philosophy of life” essential or very important?

Example 3 Modeling Options for a Toll

It’s time to decide how you’re going to pay the bridge toll so that you can get to the beach. The first option requires purchasing a transponder for $20; with the transponder, you pay a reduced toll of $3.25 each time you cross the bridge. Your second option is toll-by-plate; with this option, you pay the full toll of $4.25 each time you cross the bridge plus a $3 administrative fee for each crossing. Find the number of times you would need to cross the bridge for the costs of the two options to be the same.

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. Let

    x=the number of times you cross the bridge.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. There are no other unknown quantities, so we can skip this step.

  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. The cost with the transponder is the cost of the transponder, $20, plus the toll, $3.25, times the number of crossings, x. For toll-by-plate, you must pay both the toll, $4.25, and the administrative fee, $3, each time you cross the bridge. The cost for toll-by-plate is the sum of the toll and the administrative fee, $4.25+$3 or $7.25, times the number of crossings, x. We want to know when these two costs are equal.

    The image shows a mathematical expression 7.25x = 20 + 3.25 x
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    7.25x=20+3.25xThis is the equation that models theproblems conditions.4x=20Subtract 3.25x from both sides.4x=5Divide both sides by 4.

    Because x represents the number of times you cross the bridge, the total cost of toll-by-plate is the same as the total cost with the transponder for 5 crossings.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that the cost of toll-by-plate should be the same as the cost with the transponder. Let’s see if they are the same with 5 bridge crossings per month.

    Expression to check the proposed solution.

    If you cross the bridge five times, both options cost $36.25. Thus the proposed solution, 5 crossings, satisfies the problem’s conditions.

Check Point 3

  • You drive up to a toll plaza and find booths with attendants where you can pay the toll by cash or credit card. With this option, the toll is $5 each time you cross the bridge. The attendant gives you the option of buying a bar-coded decal for $25; with the decal, you get 25% off the normal toll of $5 for each crossing. Find the number of times you would need to cross the bridge for the costs of the two options to be the same.

Example 4 A Price Reduction on Wireless Headphones

Your favorite electronics retailer is having a terrific sale on noise cancelling wireless headphones. (You’ll finally be able to listen to music in peace!) After a 40% price reduction, you purchase headphones for $192. What was the price of the headphones before the reduction?

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the price of the headphones prior to the reduction.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. There are no other unknown quantities to find, so we can skip this step.

  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. The headphones’ original price minus the 40% reduction is the reduced price, $192.

    The image shows a mathematical expression x minus 0.4 x = 192.
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    The solution of an equation.

    The headphones’ price before the reduction was $320.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The price before the reduction, $320, minus the 40% reduction should equal the reduced price given in the original wording, $192:

    32040% of 320=3200.4(320)=320128=192.

    This verifies that the headphones’ price before the reduction was $320.

Check Point 4

  • After a 30% price reduction, you purchase a new laptop for $840. What was the laptop’s price before the reduction?

Our next example is about simple interest. Simple interest involves interest calculated only on the amount of money that we invest, called the principal. The formula I=Pr is used to find the simple interest, I, earned for one year when the principal, P, is invested at an annual interest rate, r. Dual investment problems involve different amounts of money in two or more investments, each paying a different rate.

Example 5 Solving a Dual Investment Problem

Your grandmother needs your help. She has $150,000 to invest. Part of this money is to be invested in a money market account paying 1.5% annual interest. The rest of this money is to be invested in a certificate of deposit paying 1.3% annual interest. She told you that she requires $2000 per year in extra income from the combination of these investments. How much money should be placed in each investment?

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the amount invested in the money market account at 1.5%.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. The other quantity that we seek is the amount invested at 1.3% in the certificate of deposit. Because the total amount Grandma has to invest is $150,000 and we already used up x,

    150,000x=the amount invested in the certificate of deposit at 1.3%.
  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. Because Grandma requires $2000 in total interest, the interest for the two investments combined must be $2000. Interest is Pr or rP for each investment.

    The image shows a mathematical expression 0.015 x + 0.013 left parenthesis 150,000 minus x right parenthesis = 20 00
  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    A mathematical expression steps to solve an equation to find the value of x.

    Thus,

    the amount invested at1.5%=x=25,000the amount invested at 1.3%=150,00025,000=125,000

    Grandma should invest $25,000 at 1.5% and $125,000 at 1.3%.

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The problem states that the total interest from the dual investments should be $2000. Can Grandma count on $2000 interest? The interest earned on $25,000 at 1.5% is ($25,000) (0.015), or $375. The interest earned on $125,000 at 1.3% is ($125,000)(0.013), or $1625. The total interest is $375+$1625, or $2000, exactly as it should be. You’ve made your grandmother happy. (Now if you would just visit her more often …)

Check Point 5

  • You inherited $50,000 with the stipulation that for the first year the money had to be invested in two accounts paying 0.9% and 1.1% annual interest. How much did you invest at each rate if the total interest earned for the year was $515?

Solving geometry problems usually requires a knowledge of basic geometric ideas and formulas. Formulas for area, perimeter, and volume are given in Table P.6.

Table P.6 Common Formulas for Area, Perimeter, and Volume

An illustration shows common formulas for area, perimeter, and volume of different shapes.
Table P.6 Full Alternative Text

We will be using the formula for the perimeter of a rectangle, P=2l+2w, in our next example. The formula states that a rectangle’s perimeter is the sum of twice its length and twice its width.

Example 6 Finding the Dimensions of an American Football Field

The length of an American football field is 200 feet more than the width. If the perimeter of the field is 1040 feet, what are its dimensions?

Solution

  1. Step 1. LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We know something about the length; the length is 200 feet more than the width. We will let

    x=the width.
  2. Step 2. REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. Because the length is 200 feet more than the width, we add 200 to the width to represent the length. Thus,

    x+200=the length.

    Figure P.16 illustrates an American football field and its dimensions.

    Figure P.16 An American football field

    An illustration shows an American football field with its width as x and length as x + 20 0.
  3. Step 3. WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. Because the perimeter of the field is 1040 feet,

    The image shows the mathematical expression 2 left parenthesis x + 20 0 right parenthesis + 2 x = 1040.

  4. Step 4. SOLVE THE EQUATION AND ANSWER THE QUESTION.

    Steps to solve an equation.

    Thus,

    width=x=160.length=x+200=160+200=360.

    The dimensions of an American football field are 160 feet by 360 feet. (The 360-foot length is usually described as 120 yards.)

  5. Step 5. CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. The perimeter of the football field using the dimensions that we found is

    2(160 feet)+2(360 feet)=320 feet+720 feet=1040 feet.

    Because the problem’s wording tells us that the perimeter is 1040 feet, our dimensions are correct.

Check Point 6

  • The length of a rectangular basketball court is 44 feet more than the width. If the perimeter of the basketball court is 288 feet, what are its dimensions?

We will use the formula for the area of a rectangle, A=lw, in our next example. The formula states that a rectangle’s area is the product of its length and its width.

Example 7 Solving a Problem Involving Landscape Design

A rectangular garden measures 80 feet by 60 feet. A large path of uniform width is to be added along both shorter sides and one longer side of the garden. The landscape designer doing the work wants to double the garden’s area with the addition of this path. How wide should the path be?

Solution

  1. Step 1 LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the width of the path.

    The situation is illustrated in Figure P.17. The figure shows the original 80-by-60 foot rectangular garden and the path of width x added along both shorter sides and one longer side.

    Figure P.17 The garden’s area is to be doubled by adding the path.

    An image shows a rectangular garden is surrounded by a path that has a length of 80 + 2 x and a width of 60 + x.
  2. Step 2 REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. Because the path is added along both shorter sides and one longer side, Figure P.17 shows that

     80+2x=the length of the new, expanded rectangle 60+x=the width of the new, expanded rectangle.

  3. Step 3 WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. The area of the rectangle must be doubled by the addition of the path.

    Left parenthesis 80 + 2 x right parenthesis, left parenthesis 60 + x right parenthesis = 2 times 80 times 60.
  4. Step 4 SOLVE THE EQUATION AND ANSWER THE QUESTION.

    (80+2x)(60+x)=28060This is the equation that modelsthe problems conditions.4800+200x+2x2=9600Multiply. Use FOIL on the left side.2x2+200x4800=0Subtract 9600 from both sidesand write the quadratic equationin standard form.2(x2+100x2400)=0Factor out 2, the GCF.2(x20)(x+120)=0Factor the trinomial.x20=0or x+120=0Set each variable factor equal to 0.x=20x=120Solve for x.

    The path cannot have a negative width. Because 120 is geometrically impossible, we use x=20. The width of the path should be 20 feet.

  5. Step 5 CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. Has the landscape architect doubled the garden’s area with the 20-foot-wide path? The area of the garden is 80 feet times 60 feet, or 4800 square feet. Because 80+2x and 60+x represent the length and width of the expanded rectangle,

    80+2x=80+220 =120 feet is the expanded rectangle's length.60+2x=60+20=80 feet is the expanded rectangle's width.

    The area of the expanded rectangle is 120 feet times 80 feet, or 9600 square feet. This is double the area of the garden, 4800 square feet, as specified by the problem’s conditions.

Check Point 7

  • A rectangular garden measures 16 feet by 12 feet. A path of uniform width is to be added so as to surround the entire garden. The landscape artist doing the work wants the garden and path to cover an area of 320 square feet. How wide should the path be?

In our next example, we will be using the Pythagorean Theorem to obtain a mathematical model. The ancient Greek philosopher and mathematician Pythagoras (approximately 582–500 b.c.) founded a school whose motto was “All is number.” Pythagoras is best remembered for his work with the right triangle, a triangle with one angle measuring 90°. The side opposite the 90° angle is called the hypotenuse. The other sides are called legs. Pythagoras found that if he constructed squares on each of the legs, as well as a larger square on the hypotenuse, the sum of the areas of the smaller squares is equal to the area of the larger square. This is illustrated in Figure P.18.

Figure P.18The area of the large square equals the sum of the areas of the smaller squares.

An illustration shows a right triangle of base 3, height 4, and hypotenuse 5.
Figure P. 18 Full Alternative Text

This relationship is usually stated in terms of the lengths of the three sides of a right triangle and is called the Pythagorean Theorem.

The Pythagorean Theorem

The sum of the squares of the lengths of the legs of a right triangle equals the square of the length of the hypotenuse.

If the legs have lengths a and b, and the hypotenuse has length c, then

a2+b2=c2.
An image shows a right angle triangle ABC, which represents angle C is a 90 degree angle. AC represents as a base labeled, b, Leg, BC denotes as a height labeled, a Leg, and AB is labeled, c, hypotenuse.

Example 8 Using the Pythagorean Theorem: Screen Math

A bar graph shows the rising percentage of large screen TVs on store shelves.

Source: GAP Intelligence

Did you know that the size of a television screen refers to the length of its diagonal? If the length of the HDTV screen in Figure P.19 is 28 inches and its width is 15.7 inches, what is the size of the screen to the nearest inch?

Figure P.19

A photo shows a television with a screen size of length 28 inches and a width of 15.7 inches.

Solution

Figure P.19 shows that the length, width, and diagonal of the screen form a right triangle. The diagonal is the hypotenuse of the triangle. We use the Pythagorean Theorem with a=28, b=15.7, and solve for the screen size, c.

a2+b2=c2This is the Pythagorean Theorem.282+15.72=c2Let a=28 and b=15.7.784+246.49=c2282=2828=784 and15.72=15.715.7=246.49.c2=1030.49Add: 784+246.49=1030.49. We also reversedthe two sides.
c=1030.49 or  c=1030.49Apply the square root property.c32 or  c32Use a calculator and round to thenearest inch. 1030.49    = or   1030.49 ENTER

Because c represents the size of the screen, this dimension must be positive. We reject 32. Thus, the screen size of the HDTV is 32 inches.

Check Point 8

  • Figure P.20 shows the dimensions of an old TV screen. What is the size of the screen?

    Figure P.20

    A photo shows a television with a screen size of length 25.6 inches and a width of 19.2 inches.

Example 9 Dividing the Cost of a Yacht

A group of friends agrees to share the cost of a $50,000 yacht equally. Before the purchase is made, one more person joins the group. As a result, each person’s share is reduced by $2500. How many people were in the original group?

Solution

  1. Step 1 LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We will let

    x=the number of people in the original group.
  2. Step 2 REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x.  Because one more person joined the original group, let

    x+1=the number of people in the final group.
  3. Step 3 WRITE AN EQUATION IN x THAT MODELS THE CONDITIONS. As a result of one more person joining the original group, each person’s share is reduced by $2500.

    The image shows a mathematical expression, start fraction 50,000 over x end fraction minus 2500 = start fraction 50,000 over x + 1 end fraction.
  4. Step 4 SOLVE THE EQUATION AND ANSWER THE QUESTION.

    50,000x2500=50,000x+1This is the equation that models theproblems conditions.x(x+1)(50,000x2500)=x(x+1)50,000x+1Multiply both sides by x(x+1),the LCD. x (x+1)50,000 x x(x+1)2500=x (x+1) 50,000 (x+1) Use the distributive property anddivide out common factors.50,000(x+1)2500x(x+1)=50,000xSimplify.50,000x+50,0002500x22500x=50,000xUse the distributive property.2500x2+47,500x+50,000=50,000xCombine like terms.2500x22500x+50,000=0Write the quadratic equation instandard form.2500(x2+x20)=0Factor out 2500.2500(x+5)(x4)=0Factor completely.x+5=0or x4=0Set each variable factor equal to zero.x=5x=4Solve the resulting equations.

    Because x represents the number of people in the original group, x cannot be negative. Thus, there were four people in the original group.

  5. Step 5 CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM.

     original cost per person=$50,0004=$12,500 final cost per person=$50,0005=$10,000

    We see that each person’s share is reduced by $12,500$10,000, or $2500, as specified by the problem’s conditions.

Check Point 9

  • A group of people share equally in a $5,000,000 lottery. Before the money is divided, three more winning ticket holders are declared. As a result, each person’s share is reduced by $375,000. How many people were in the original group of winners?

P.8: Exercise Set

P.8 Exercise Set

Practice and Application Exercises

How will you spend your average life expectancy of 78 years? The bar graph shows the average number of years you will devote to each of your most time-consuming activities. Exercises 12 are based on the data displayed by the graph.

The image shows a bar graph titled, How You Will Spend Your Average Life Expectancy of 78 Years.

Source: U.S. Bureau of Labor Statistics

  1. 1. According to the U.S. Bureau of Labor Statistics, you will devote 37 years to sleeping and watching TV. The number of years sleeping will exceed the number of years watching TV by 19. Over your lifetime, how many years will you spend on each of these activities?

  2. 2. According to the U.S. Bureau of Labor Statistics, you will devote 32 years to sleeping and eating. The number of years sleeping will exceed the number of years eating by 24. Over your lifetime, how many years will you spend on each of these activities?

The bar graph shows median yearly earnings of full-time workers in the United States for people 25 years and over for three occupations and two levels of education. Exercises 34 are based on the data displayed by the graph.

A bar graph titled, Median earning of full- time U.S. workers 25 years and over by occupation and level of education.

Sources: U.S. Census Bureau; Education Pays 2019

  1. 3. The median yearly salary of a general manager with a bachelor’s degree or higher is $31,000 less than twice that of a general manager with just a high school diploma. Combined, two managers with each of these educational attainments earn $149,300. Find the median yearly salary of general managers with each of these levels of education.

  2. 4. The median yearly salary of a retail salesperson with a bachelor’s degree or higher is $14,300 less than twice that of a retail salesperson with just a high school diploma. Combined, two salespeople with each of these educational attainments earn $79,900. Find the median yearly salary of salespeople with each of these levels of education.

Despite booming new car sales with their cha-ching sounds, the average age of vehicles on U.S. roads is not going down. The bar graph shows the average price of new cars in the United States and the average age of cars on U.S. roads for two selected years. Exercises 56 are based on the information displayed by the graph.

A bar graph titled, Average price of new cars and Average age of cars on U.S. roads.

Source: Kelley Blue Book, IHS Automotive/Polk

  1. 5. In 2019, the average price of a new car was $38,900. For the period shown, new-car prices increased by approximately $800 per year. If this trend continues, how many years after 2019 will the price of a new car average $44,500? In which year will this occur?

  2. 6. In 2019, the average age of cars on U.S. roads was 11.8 years. For the period shown, this average age increased by approximately 0.15 year per year. If this trend continues, how many years after 2019 will the average age of vehicles on U.S. roads be 13 years? In which year will this occur?

  3. 7. A new car worth $36,000 is depreciating in value by $4000 per year.

    1. Write a formula that models the car’s value, y, in dollars, after x years.

    2. Use the formula from part (a) to determine after how many years the car’s value will be $12,000.

  4. 8. A new car worth $45,000 is depreciating in value by $5000 per year.

    1. Write a formula that models the car’s value, y, in dollars, after x years.

    2. Use the formula from part (a) to determine after how many years the car’s value will be $10,000.

  5. 9. You are choosing between two gyms. One gym offers membership for a fee of $40 plus a monthly fee of $25. The other offers membership for a fee of $15 plus a monthly fee of $30. After how many months will the total cost at each gym be the same? What will be the total cost for each gym?

  6. 10. Taxi rates are determined by local authorities. In New York City, the night-time cost of a taxi includes a base fee of $3 plus a charge of $1.56 per kilometer. In Boston, regardless of the time of day, the base fee is $2.60 with a charge of $1.75 per kilometer. For how many kilometers will the cost of a night-time taxi ride in each city be the same? Round to the nearest kilometer. What will the cost be in each city for the rounded number of kilometers?

  7. 11. A transponder for a toll bridge costs $27.50. With the transponder, the toll is $5 each time you cross the bridge. The only other option is toll-by-plate, for which the toll is $6.25 each time you cross the bridge with an additional administrative fee of $1.25 for each crossing. How many times would you need to cross the bridge for the costs of the two toll options to be the same?

  8. 12. An electronic pass for a toll road costs $30. The toll is normally $5.00 but is reduced by 30% for people who have purchased the electronic pass. Determine the number of times the road must be used so that the total cost without the pass is the same as the total cost with the pass.

  9. 13. In 2010, there were 13,300 students at college A, with a projected enrollment increase of 1000 students per year. In the same year, there were 26,800 students at college B, with a projected enrollment decline of 500 students per year. According to these projections, when did the colleges have the same enrollment? What was the enrollment in each college at that time?

  10. 14. In 2000, the population of Greece was 10,600,000, with projections of a population decrease of 28,000 people per year. In the same year, the population of Belgium was 10,200,000, with projections of a population decrease of 12,000 people per year. (Source: United Nations) According to these projections, when will the two countries have the same population? What will be the population at that time?

  11. 15. After a 20% reduction, you purchase a television for $336. What was the television’s price before the reduction?

  12. 16. After a 30% reduction, you purchase wireless earbuds for $90.30. What was the earbuds’ price before the reduction?

  13. 17. Including a 10.5% hotel tax, your room in San Diego cost $216.58 per night. Find the nightly cost before the tax was added.

  14. 18. Including a 17.4% hotel tax, your room in Chicago cost $287.63 per night. Find the nightly cost before the tax was added.

Exercises 1920 involve markup, the amount added to the dealer’s cost of an item to arrive at the selling price of that item.

  1. 19. The selling price of a refrigerator is $1198. If the markup is 25% of the dealer’s cost, what is the dealer’s cost of the refrigerator?

  2. 20. The selling price of a scientific calculator is $15. If the markup is 25% of the dealer’s cost, what is the dealer’s cost of the calculator?

  3. 21. You invested $20,000 in two accounts paying 1.45% and 1.59% annual interest. If the total interest earned for the year was $307.50, how much was invested at each rate?

  4. 22. You invested $30,000 in two accounts paying 2.19% and 2.45% annual interest. If the total interest earned for the year was $705.88, how much was invested at each rate?

  5. 23. Things did not go quite as planned. You invested $10,000, part of it in a stock that realized a 12% gain. However, the rest of the money suffered a 5% loss. If you had an overall gain of $520, how much was invested at each rate?

  6. 24. Things did not go quite as planned. You invested $15,000, part of it in a stock that realized a 15% gain. However, the rest of the money suffered a 7% loss. If you had an overall gain of $1590, how much was invested at each rate?

  7. 25. A rectangular soccer field is twice as long as it is wide. If the perimeter of the soccer field is 300 yards, what are its dimensions?

  8. 26. A rectangular swimming pool is three times as long as it is wide. If the perimeter of the pool is 320 feet, what are its dimensions?

  9. 27. The length of the rectangular tennis court at Wimbledon is 6 feet longer than twice the width. If the court’s perimeter is 228 feet, what are the court’s dimensions?

  10. 28. The length of a rectangular pool is 6 meters less than twice the width. If the pool’s perimeter is 126 meters, what are its dimensions?

  11. 29. The rectangular painting in the figure shown measures 12 inches by 16 inches and is surrounded by a frame of uniform width around the four edges. The perimeter of the rectangle formed by the painting and its frame is 72 inches. Determine the width of the frame.

    The image shows a rectangular painting of length 16 inches and width 12 inches and is surrounded by a frame of uniform width around the four edges labeled x.
  12. 30. The rectangular swimming pool in the figure shown measures 40 feet by 60 feet and is surrounded by a path of uniform width around the four edges. The perimeter of the rectangle formed by the pool and the surrounding path is 248 feet. Determine the width of the path.

    The image shows the rectangular swimming pool measures length of 60 feet and width of 40 feet and is surrounded by a path of uniform width around the four edges labeled x.

  13. 31. The length of a rectangular sign is 3 feet longer than the width. If the sign’s area is 54 square feet, find its length and width.

  14. 32. A rectangular parking lot has a length that is 3 yards greater than the width. The area of the parking lot is 180 square yards. Find the length and the width.

  15. 33. Each side of a square is lengthened by 3 inches. The area of this new, larger square is 64 square inches. Find the length of a side of the original square.

  16. 34. Each side of a square is lengthened by 2 inches. The area of this new, larger square is 36 square inches. Find the length of a side of the original square.

  17. 35. A pool measuring 10 meters by 20 meters is surrounded by a path of uniform width. If the area of the pool and the path combined is 600 square meters, what is the width of the path?

  18. 36. A vacant rectangular lot is being turned into a community vegetable garden measuring 15 meters by 12 meters. A path of uniform width is to surround the garden. If the area of the lot is 378 square meters, find the width of the path surrounding the garden.

  19. 37. As part of a landscaping project, you put in a flower bed measuring 20 feet by 30 feet. To finish off the project, you are putting in a uniform border of pine bark around the outside of the rectangular garden. You have enough pine bark to cover 336 square feet. How wide should the border be?

  20. 38. As part of a landscaping project, you put in a flower bed measuring 10 feet by 12 feet. You plan to surround the bed with a uniform border of low-growing plants that require 1 square foot each when mature. If you have 168 of these plants, how wide a strip around the flower bed should you prepare for the border?

  21. 39. A 20-foot ladder is 15 feet from a house. How far up the house, to the nearest tenth of a foot, does the ladder reach?

  22. 40. The base of a 30-foot ladder is 10 feet from a building. If the ladder reaches the flat roof, how tall, to the nearest tenth of a foot, is the building?

  23. 41. A tree is supported by a wire anchored in the ground 5 feet from its base. The wire is 1 foot longer than the height that it reaches on the tree. Find the length of the wire.

  24. 42. A tree is supported by a wire anchored in the ground 15 feet from its base. The wire is 4 feet longer than the height that it reaches on the tree. Find the length of the wire.

  25. 43. A rectangular piece of land whose length its twice its width has a diagonal distance of 64 yards. How many yards, to the nearest tenth of a yard, does a person save by walking diagonally across the land instead of walking its length and its width?

  26. 44. A rectangular piece of land whose length is three times its width has a diagonal distance of 92 yards. How many yards, to the nearest tenth of a yard, does a person save by walking diagonally across the land instead of walking its length and its width?

  27. 45. A group of people share equally in a $20,000,000 lottery. Before the money is divided, two more winning ticket holders are declared. As a result, each person’s share is reduced by $500,000. How many people were in the original group of winners?

  28. 46. A group of friends agrees to share the cost of a $480,000 vacation condominium equally. Before the purchase is made, four more people join the group and enter the agreement. As a result, each person’s share is reduced by $32,000. How many people were in the original group?

In Exercises 4750, use the formula

Time traveled=Distance traveledAverage velocity.
  1. 47. A car can travel 300 miles in the same amount of time it takes a bus to travel 180 miles. If the average velocity of the bus is 20 miles per hour slower than the average velocity of the car, find the average velocity for each.

  2. 48. A passenger train can travel 240 miles in the same amount of time it takes a freight train to travel 160 miles. If the average velocity of the freight train is 20 miles per hour slower than the average velocity of the passenger train, find the average velocity of each.

  3. 49. You ride your bike to campus a distance of 5 miles and return home on the same route. Going to campus, you ride mostly downhill and average 9 miles per hour faster than on your return trip home. If the round trip takes one hour and ten minutes—that is 76 hours—what is your average velocity on the return trip?

  4. 50. An engine pulls a train 140 miles. Then a second engine, whose average velocity is 5 miles per hour faster than the first engine, takes over and pulls the train 200 miles. The total time required for both engines is 9 hours. Find the average velocity of each engine.

  5. 51. An automobile repair shop charged a customer $1182, listing $357 for parts and the remainder for labor. If the cost of labor is $75 per hour, how many hours of labor did it take to repair the car?

  6. 52. A repair bill on a sailboat came to $2356, including $826 for parts and the remainder for labor. If the cost of labor is $90 per hour, how many hours of labor did it take to repair the sailboat?

  7. 53. For an international telephone call, a telephone company charges $0.43 for the first minute, $0.32 for each additional minute, and a $2.10 service charge. If the cost of a call is $5.73, how long did the person talk?

  8. 54. A job pays an annual salary of $57,900, which includes a holiday bonus of $1500. If paychecks are issued twice a month, what is the gross amount for each paycheck?

  9. 55. You have 35 hits in 140 times at bat. Your batting average is 35140, or 0.25. How many consecutive hits must you get to increase your batting average to 0.30?

  10. 56. You have 30 hits in 120 times at bat. Your batting average is 30120, or 0.25. How many consecutive hits must you get to increase your batting average to 0.28?

Explaining the Concepts

  1. 57. In your own words, describe a step-by-step approach for solving algebraic word problems.

  2. 58. Write an original word problem that can be solved using a linear equation. Then solve the problem.

Critical Thinking Exercises

MAKE SENSE? In Exercises 5961, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 59. By modeling attitudes of college freshmen from 1969 through 2018, I can make precise predictions about the attitudes of the freshman class of 2040.

  2. 60. I find the hardest part in solving a word problem is writing the equation that models the verbal conditions.

  3. 61. After a 35% reduction, a laptop’s price is $780, so I determined the original price, x, by solving x0.35=780.

  4. 62. At the north campus of a performing arts school, 10% of the students are music majors. At the south campus, 90% of the students are music majors. The campuses are merged into one east campus. If 42% of the 1000 students at the east campus are music majors, how many students did the north and south campuses have before the merger?

  5. 63. The price of a dress is reduced by 40%. When the dress still does not sell, it is reduced by 40% of the reduced price. If the price of the dress after both reductions is $72, what was the original price?

  6. 64. If I am three times your age and in 20 years I’ll be twice your age, how old are we?

  7. 65. Suppose that we agree to pay you 8¢ for every problem in this chapter that you solve correctly and fine you 5¢ for every problem done incorrectly. If at the end of 26 problems we do not owe each other any money, how many problems did you solve correctly?

  8. 66. It was wartime when Dick and Jane found out Jane was pregnant. Dick was drafted and made out a will, deciding that $14,000 in a savings account was to be divided between his wife and his child-to-be. Rather strangely, and certainly with gender bias, Dick stipulated that if the child were a boy, he would get twice the amount of the mother’s portion. If it were a girl, the mother would get twice the amount the girl was to receive. We’ll never know what Dick was thinking of, for (as fate would have it) he did not return from war. Jane gave birth to twins—a boy and a girl. How was the money divided?

  9. 67. A thief steals a number of rare plants from a nursery. On the way out, the thief meets three security guards, one after another. To each security guard, the thief is forced to give one-half the plants that he still has, plus two more. Finally, the thief leaves the nursery with one lone palm. How many plants were originally stolen?

Group Exercise

  1. 68. One of the best ways to learn how to solve a word problem in algebra is to design word problems of your own. Creating a word problem makes you very aware of precisely how much information is needed to solve the problem. You must also focus on the best way to present information to a reader and on how much information to give. As you write your problem, you gain skills that will help you solve problems created by others.

    The group should design five different word problems that can be solved using linear equations. All of the problems should be on different topics. For example, the group should not have more than one problem on simple interest. The group should turn in both the problems and their algebraic solutions.

Preview Exercises

Exercises 6971 will help you prepare for the material covered in the next section.

  1. 69. Is 1 a solution of 32x11?

  2. 70. Solve: 2x4=x+5.

  3. 71. Solve: x+34=x23+14.

P.8: Exercise Set

P.8 Exercise Set

Practice and Application Exercises

How will you spend your average life expectancy of 78 years? The bar graph shows the average number of years you will devote to each of your most time-consuming activities. Exercises 12 are based on the data displayed by the graph.

The image shows a bar graph titled, How You Will Spend Your Average Life Expectancy of 78 Years.

Source: U.S. Bureau of Labor Statistics

  1. 1. According to the U.S. Bureau of Labor Statistics, you will devote 37 years to sleeping and watching TV. The number of years sleeping will exceed the number of years watching TV by 19. Over your lifetime, how many years will you spend on each of these activities?

  2. 2. According to the U.S. Bureau of Labor Statistics, you will devote 32 years to sleeping and eating. The number of years sleeping will exceed the number of years eating by 24. Over your lifetime, how many years will you spend on each of these activities?

The bar graph shows median yearly earnings of full-time workers in the United States for people 25 years and over for three occupations and two levels of education. Exercises 34 are based on the data displayed by the graph.

A bar graph titled, Median earning of full- time U.S. workers 25 years and over by occupation and level of education.

Sources: U.S. Census Bureau; Education Pays 2019

  1. 3. The median yearly salary of a general manager with a bachelor’s degree or higher is $31,000 less than twice that of a general manager with just a high school diploma. Combined, two managers with each of these educational attainments earn $149,300. Find the median yearly salary of general managers with each of these levels of education.

  2. 4. The median yearly salary of a retail salesperson with a bachelor’s degree or higher is $14,300 less than twice that of a retail salesperson with just a high school diploma. Combined, two salespeople with each of these educational attainments earn $79,900. Find the median yearly salary of salespeople with each of these levels of education.

Despite booming new car sales with their cha-ching sounds, the average age of vehicles on U.S. roads is not going down. The bar graph shows the average price of new cars in the United States and the average age of cars on U.S. roads for two selected years. Exercises 56 are based on the information displayed by the graph.

A bar graph titled, Average price of new cars and Average age of cars on U.S. roads.

Source: Kelley Blue Book, IHS Automotive/Polk

  1. 5. In 2019, the average price of a new car was $38,900. For the period shown, new-car prices increased by approximately $800 per year. If this trend continues, how many years after 2019 will the price of a new car average $44,500? In which year will this occur?

  2. 6. In 2019, the average age of cars on U.S. roads was 11.8 years. For the period shown, this average age increased by approximately 0.15 year per year. If this trend continues, how many years after 2019 will the average age of vehicles on U.S. roads be 13 years? In which year will this occur?

  3. 7. A new car worth $36,000 is depreciating in value by $4000 per year.

    1. Write a formula that models the car’s value, y, in dollars, after x years.

    2. Use the formula from part (a) to determine after how many years the car’s value will be $12,000.

  4. 8. A new car worth $45,000 is depreciating in value by $5000 per year.

    1. Write a formula that models the car’s value, y, in dollars, after x years.

    2. Use the formula from part (a) to determine after how many years the car’s value will be $10,000.

  5. 9. You are choosing between two gyms. One gym offers membership for a fee of $40 plus a monthly fee of $25. The other offers membership for a fee of $15 plus a monthly fee of $30. After how many months will the total cost at each gym be the same? What will be the total cost for each gym?

  6. 10. Taxi rates are determined by local authorities. In New York City, the night-time cost of a taxi includes a base fee of $3 plus a charge of $1.56 per kilometer. In Boston, regardless of the time of day, the base fee is $2.60 with a charge of $1.75 per kilometer. For how many kilometers will the cost of a night-time taxi ride in each city be the same? Round to the nearest kilometer. What will the cost be in each city for the rounded number of kilometers?

  7. 11. A transponder for a toll bridge costs $27.50. With the transponder, the toll is $5 each time you cross the bridge. The only other option is toll-by-plate, for which the toll is $6.25 each time you cross the bridge with an additional administrative fee of $1.25 for each crossing. How many times would you need to cross the bridge for the costs of the two toll options to be the same?

  8. 12. An electronic pass for a toll road costs $30. The toll is normally $5.00 but is reduced by 30% for people who have purchased the electronic pass. Determine the number of times the road must be used so that the total cost without the pass is the same as the total cost with the pass.

  9. 13. In 2010, there were 13,300 students at college A, with a projected enrollment increase of 1000 students per year. In the same year, there were 26,800 students at college B, with a projected enrollment decline of 500 students per year. According to these projections, when did the colleges have the same enrollment? What was the enrollment in each college at that time?

  10. 14. In 2000, the population of Greece was 10,600,000, with projections of a population decrease of 28,000 people per year. In the same year, the population of Belgium was 10,200,000, with projections of a population decrease of 12,000 people per year. (Source: United Nations) According to these projections, when will the two countries have the same population? What will be the population at that time?

  11. 15. After a 20% reduction, you purchase a television for $336. What was the television’s price before the reduction?

  12. 16. After a 30% reduction, you purchase wireless earbuds for $90.30. What was the earbuds’ price before the reduction?

  13. 17. Including a 10.5% hotel tax, your room in San Diego cost $216.58 per night. Find the nightly cost before the tax was added.

  14. 18. Including a 17.4% hotel tax, your room in Chicago cost $287.63 per night. Find the nightly cost before the tax was added.

Exercises 1920 involve markup, the amount added to the dealer’s cost of an item to arrive at the selling price of that item.

  1. 19. The selling price of a refrigerator is $1198. If the markup is 25% of the dealer’s cost, what is the dealer’s cost of the refrigerator?

  2. 20. The selling price of a scientific calculator is $15. If the markup is 25% of the dealer’s cost, what is the dealer’s cost of the calculator?

  3. 21. You invested $20,000 in two accounts paying 1.45% and 1.59% annual interest. If the total interest earned for the year was $307.50, how much was invested at each rate?

  4. 22. You invested $30,000 in two accounts paying 2.19% and 2.45% annual interest. If the total interest earned for the year was $705.88, how much was invested at each rate?

  5. 23. Things did not go quite as planned. You invested $10,000, part of it in a stock that realized a 12% gain. However, the rest of the money suffered a 5% loss. If you had an overall gain of $520, how much was invested at each rate?

  6. 24. Things did not go quite as planned. You invested $15,000, part of it in a stock that realized a 15% gain. However, the rest of the money suffered a 7% loss. If you had an overall gain of $1590, how much was invested at each rate?

  7. 25. A rectangular soccer field is twice as long as it is wide. If the perimeter of the soccer field is 300 yards, what are its dimensions?

  8. 26. A rectangular swimming pool is three times as long as it is wide. If the perimeter of the pool is 320 feet, what are its dimensions?

  9. 27. The length of the rectangular tennis court at Wimbledon is 6 feet longer than twice the width. If the court’s perimeter is 228 feet, what are the court’s dimensions?

  10. 28. The length of a rectangular pool is 6 meters less than twice the width. If the pool’s perimeter is 126 meters, what are its dimensions?

  11. 29. The rectangular painting in the figure shown measures 12 inches by 16 inches and is surrounded by a frame of uniform width around the four edges. The perimeter of the rectangle formed by the painting and its frame is 72 inches. Determine the width of the frame.

    The image shows a rectangular painting of length 16 inches and width 12 inches and is surrounded by a frame of uniform width around the four edges labeled x.
  12. 30. The rectangular swimming pool in the figure shown measures 40 feet by 60 feet and is surrounded by a path of uniform width around the four edges. The perimeter of the rectangle formed by the pool and the surrounding path is 248 feet. Determine the width of the path.

    The image shows the rectangular swimming pool measures length of 60 feet and width of 40 feet and is surrounded by a path of uniform width around the four edges labeled x.

  13. 31. The length of a rectangular sign is 3 feet longer than the width. If the sign’s area is 54 square feet, find its length and width.

  14. 32. A rectangular parking lot has a length that is 3 yards greater than the width. The area of the parking lot is 180 square yards. Find the length and the width.

  15. 33. Each side of a square is lengthened by 3 inches. The area of this new, larger square is 64 square inches. Find the length of a side of the original square.

  16. 34. Each side of a square is lengthened by 2 inches. The area of this new, larger square is 36 square inches. Find the length of a side of the original square.

  17. 35. A pool measuring 10 meters by 20 meters is surrounded by a path of uniform width. If the area of the pool and the path combined is 600 square meters, what is the width of the path?

  18. 36. A vacant rectangular lot is being turned into a community vegetable garden measuring 15 meters by 12 meters. A path of uniform width is to surround the garden. If the area of the lot is 378 square meters, find the width of the path surrounding the garden.

  19. 37. As part of a landscaping project, you put in a flower bed measuring 20 feet by 30 feet. To finish off the project, you are putting in a uniform border of pine bark around the outside of the rectangular garden. You have enough pine bark to cover 336 square feet. How wide should the border be?

  20. 38. As part of a landscaping project, you put in a flower bed measuring 10 feet by 12 feet. You plan to surround the bed with a uniform border of low-growing plants that require 1 square foot each when mature. If you have 168 of these plants, how wide a strip around the flower bed should you prepare for the border?

  21. 39. A 20-foot ladder is 15 feet from a house. How far up the house, to the nearest tenth of a foot, does the ladder reach?

  22. 40. The base of a 30-foot ladder is 10 feet from a building. If the ladder reaches the flat roof, how tall, to the nearest tenth of a foot, is the building?

  23. 41. A tree is supported by a wire anchored in the ground 5 feet from its base. The wire is 1 foot longer than the height that it reaches on the tree. Find the length of the wire.

  24. 42. A tree is supported by a wire anchored in the ground 15 feet from its base. The wire is 4 feet longer than the height that it reaches on the tree. Find the length of the wire.

  25. 43. A rectangular piece of land whose length its twice its width has a diagonal distance of 64 yards. How many yards, to the nearest tenth of a yard, does a person save by walking diagonally across the land instead of walking its length and its width?

  26. 44. A rectangular piece of land whose length is three times its width has a diagonal distance of 92 yards. How many yards, to the nearest tenth of a yard, does a person save by walking diagonally across the land instead of walking its length and its width?

  27. 45. A group of people share equally in a $20,000,000 lottery. Before the money is divided, two more winning ticket holders are declared. As a result, each person’s share is reduced by $500,000. How many people were in the original group of winners?

  28. 46. A group of friends agrees to share the cost of a $480,000 vacation condominium equally. Before the purchase is made, four more people join the group and enter the agreement. As a result, each person’s share is reduced by $32,000. How many people were in the original group?

In Exercises 4750, use the formula

Time traveled=Distance traveledAverage velocity.
  1. 47. A car can travel 300 miles in the same amount of time it takes a bus to travel 180 miles. If the average velocity of the bus is 20 miles per hour slower than the average velocity of the car, find the average velocity for each.

  2. 48. A passenger train can travel 240 miles in the same amount of time it takes a freight train to travel 160 miles. If the average velocity of the freight train is 20 miles per hour slower than the average velocity of the passenger train, find the average velocity of each.

  3. 49. You ride your bike to campus a distance of 5 miles and return home on the same route. Going to campus, you ride mostly downhill and average 9 miles per hour faster than on your return trip home. If the round trip takes one hour and ten minutes—that is 76 hours—what is your average velocity on the return trip?

  4. 50. An engine pulls a train 140 miles. Then a second engine, whose average velocity is 5 miles per hour faster than the first engine, takes over and pulls the train 200 miles. The total time required for both engines is 9 hours. Find the average velocity of each engine.

  5. 51. An automobile repair shop charged a customer $1182, listing $357 for parts and the remainder for labor. If the cost of labor is $75 per hour, how many hours of labor did it take to repair the car?

  6. 52. A repair bill on a sailboat came to $2356, including $826 for parts and the remainder for labor. If the cost of labor is $90 per hour, how many hours of labor did it take to repair the sailboat?

  7. 53. For an international telephone call, a telephone company charges $0.43 for the first minute, $0.32 for each additional minute, and a $2.10 service charge. If the cost of a call is $5.73, how long did the person talk?

  8. 54. A job pays an annual salary of $57,900, which includes a holiday bonus of $1500. If paychecks are issued twice a month, what is the gross amount for each paycheck?

  9. 55. You have 35 hits in 140 times at bat. Your batting average is 35140, or 0.25. How many consecutive hits must you get to increase your batting average to 0.30?

  10. 56. You have 30 hits in 120 times at bat. Your batting average is 30120, or 0.25. How many consecutive hits must you get to increase your batting average to 0.28?

Explaining the Concepts

  1. 57. In your own words, describe a step-by-step approach for solving algebraic word problems.

  2. 58. Write an original word problem that can be solved using a linear equation. Then solve the problem.

Critical Thinking Exercises

MAKE SENSE? In Exercises 5961, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 59. By modeling attitudes of college freshmen from 1969 through 2018, I can make precise predictions about the attitudes of the freshman class of 2040.

  2. 60. I find the hardest part in solving a word problem is writing the equation that models the verbal conditions.

  3. 61. After a 35% reduction, a laptop’s price is $780, so I determined the original price, x, by solving x0.35=780.

  4. 62. At the north campus of a performing arts school, 10% of the students are music majors. At the south campus, 90% of the students are music majors. The campuses are merged into one east campus. If 42% of the 1000 students at the east campus are music majors, how many students did the north and south campuses have before the merger?

  5. 63. The price of a dress is reduced by 40%. When the dress still does not sell, it is reduced by 40% of the reduced price. If the price of the dress after both reductions is $72, what was the original price?

  6. 64. If I am three times your age and in 20 years I’ll be twice your age, how old are we?

  7. 65. Suppose that we agree to pay you 8¢ for every problem in this chapter that you solve correctly and fine you 5¢ for every problem done incorrectly. If at the end of 26 problems we do not owe each other any money, how many problems did you solve correctly?

  8. 66. It was wartime when Dick and Jane found out Jane was pregnant. Dick was drafted and made out a will, deciding that $14,000 in a savings account was to be divided between his wife and his child-to-be. Rather strangely, and certainly with gender bias, Dick stipulated that if the child were a boy, he would get twice the amount of the mother’s portion. If it were a girl, the mother would get twice the amount the girl was to receive. We’ll never know what Dick was thinking of, for (as fate would have it) he did not return from war. Jane gave birth to twins—a boy and a girl. How was the money divided?

  9. 67. A thief steals a number of rare plants from a nursery. On the way out, the thief meets three security guards, one after another. To each security guard, the thief is forced to give one-half the plants that he still has, plus two more. Finally, the thief leaves the nursery with one lone palm. How many plants were originally stolen?

Group Exercise

  1. 68. One of the best ways to learn how to solve a word problem in algebra is to design word problems of your own. Creating a word problem makes you very aware of precisely how much information is needed to solve the problem. You must also focus on the best way to present information to a reader and on how much information to give. As you write your problem, you gain skills that will help you solve problems created by others.

    The group should design five different word problems that can be solved using linear equations. All of the problems should be on different topics. For example, the group should not have more than one problem on simple interest. The group should turn in both the problems and their algebraic solutions.

Preview Exercises

Exercises 6971 will help you prepare for the material covered in the next section.

  1. 69. Is 1 a solution of 32x11?

  2. 70. Solve: 2x4=x+5.

  3. 71. Solve: x+34=x23+14.

P.8: Exercise Set

P.8 Exercise Set

Practice and Application Exercises

How will you spend your average life expectancy of 78 years? The bar graph shows the average number of years you will devote to each of your most time-consuming activities. Exercises 12 are based on the data displayed by the graph.

The image shows a bar graph titled, How You Will Spend Your Average Life Expectancy of 78 Years.

Source: U.S. Bureau of Labor Statistics

  1. 1. According to the U.S. Bureau of Labor Statistics, you will devote 37 years to sleeping and watching TV. The number of years sleeping will exceed the number of years watching TV by 19. Over your lifetime, how many years will you spend on each of these activities?

  2. 2. According to the U.S. Bureau of Labor Statistics, you will devote 32 years to sleeping and eating. The number of years sleeping will exceed the number of years eating by 24. Over your lifetime, how many years will you spend on each of these activities?

The bar graph shows median yearly earnings of full-time workers in the United States for people 25 years and over for three occupations and two levels of education. Exercises 34 are based on the data displayed by the graph.

A bar graph titled, Median earning of full- time U.S. workers 25 years and over by occupation and level of education.

Sources: U.S. Census Bureau; Education Pays 2019

  1. 3. The median yearly salary of a general manager with a bachelor’s degree or higher is $31,000 less than twice that of a general manager with just a high school diploma. Combined, two managers with each of these educational attainments earn $149,300. Find the median yearly salary of general managers with each of these levels of education.

  2. 4. The median yearly salary of a retail salesperson with a bachelor’s degree or higher is $14,300 less than twice that of a retail salesperson with just a high school diploma. Combined, two salespeople with each of these educational attainments earn $79,900. Find the median yearly salary of salespeople with each of these levels of education.

Despite booming new car sales with their cha-ching sounds, the average age of vehicles on U.S. roads is not going down. The bar graph shows the average price of new cars in the United States and the average age of cars on U.S. roads for two selected years. Exercises 56 are based on the information displayed by the graph.

A bar graph titled, Average price of new cars and Average age of cars on U.S. roads.

Source: Kelley Blue Book, IHS Automotive/Polk

  1. 5. In 2019, the average price of a new car was $38,900. For the period shown, new-car prices increased by approximately $800 per year. If this trend continues, how many years after 2019 will the price of a new car average $44,500? In which year will this occur?

  2. 6. In 2019, the average age of cars on U.S. roads was 11.8 years. For the period shown, this average age increased by approximately 0.15 year per year. If this trend continues, how many years after 2019 will the average age of vehicles on U.S. roads be 13 years? In which year will this occur?

  3. 7. A new car worth $36,000 is depreciating in value by $4000 per year.

    1. Write a formula that models the car’s value, y, in dollars, after x years.

    2. Use the formula from part (a) to determine after how many years the car’s value will be $12,000.

  4. 8. A new car worth $45,000 is depreciating in value by $5000 per year.

    1. Write a formula that models the car’s value, y, in dollars, after x years.

    2. Use the formula from part (a) to determine after how many years the car’s value will be $10,000.

  5. 9. You are choosing between two gyms. One gym offers membership for a fee of $40 plus a monthly fee of $25. The other offers membership for a fee of $15 plus a monthly fee of $30. After how many months will the total cost at each gym be the same? What will be the total cost for each gym?

  6. 10. Taxi rates are determined by local authorities. In New York City, the night-time cost of a taxi includes a base fee of $3 plus a charge of $1.56 per kilometer. In Boston, regardless of the time of day, the base fee is $2.60 with a charge of $1.75 per kilometer. For how many kilometers will the cost of a night-time taxi ride in each city be the same? Round to the nearest kilometer. What will the cost be in each city for the rounded number of kilometers?

  7. 11. A transponder for a toll bridge costs $27.50. With the transponder, the toll is $5 each time you cross the bridge. The only other option is toll-by-plate, for which the toll is $6.25 each time you cross the bridge with an additional administrative fee of $1.25 for each crossing. How many times would you need to cross the bridge for the costs of the two toll options to be the same?

  8. 12. An electronic pass for a toll road costs $30. The toll is normally $5.00 but is reduced by 30% for people who have purchased the electronic pass. Determine the number of times the road must be used so that the total cost without the pass is the same as the total cost with the pass.

  9. 13. In 2010, there were 13,300 students at college A, with a projected enrollment increase of 1000 students per year. In the same year, there were 26,800 students at college B, with a projected enrollment decline of 500 students per year. According to these projections, when did the colleges have the same enrollment? What was the enrollment in each college at that time?

  10. 14. In 2000, the population of Greece was 10,600,000, with projections of a population decrease of 28,000 people per year. In the same year, the population of Belgium was 10,200,000, with projections of a population decrease of 12,000 people per year. (Source: United Nations) According to these projections, when will the two countries have the same population? What will be the population at that time?

  11. 15. After a 20% reduction, you purchase a television for $336. What was the television’s price before the reduction?

  12. 16. After a 30% reduction, you purchase wireless earbuds for $90.30. What was the earbuds’ price before the reduction?

  13. 17. Including a 10.5% hotel tax, your room in San Diego cost $216.58 per night. Find the nightly cost before the tax was added.

  14. 18. Including a 17.4% hotel tax, your room in Chicago cost $287.63 per night. Find the nightly cost before the tax was added.

Exercises 1920 involve markup, the amount added to the dealer’s cost of an item to arrive at the selling price of that item.

  1. 19. The selling price of a refrigerator is $1198. If the markup is 25% of the dealer’s cost, what is the dealer’s cost of the refrigerator?

  2. 20. The selling price of a scientific calculator is $15. If the markup is 25% of the dealer’s cost, what is the dealer’s cost of the calculator?

  3. 21. You invested $20,000 in two accounts paying 1.45% and 1.59% annual interest. If the total interest earned for the year was $307.50, how much was invested at each rate?

  4. 22. You invested $30,000 in two accounts paying 2.19% and 2.45% annual interest. If the total interest earned for the year was $705.88, how much was invested at each rate?

  5. 23. Things did not go quite as planned. You invested $10,000, part of it in a stock that realized a 12% gain. However, the rest of the money suffered a 5% loss. If you had an overall gain of $520, how much was invested at each rate?

  6. 24. Things did not go quite as planned. You invested $15,000, part of it in a stock that realized a 15% gain. However, the rest of the money suffered a 7% loss. If you had an overall gain of $1590, how much was invested at each rate?

  7. 25. A rectangular soccer field is twice as long as it is wide. If the perimeter of the soccer field is 300 yards, what are its dimensions?

  8. 26. A rectangular swimming pool is three times as long as it is wide. If the perimeter of the pool is 320 feet, what are its dimensions?

  9. 27. The length of the rectangular tennis court at Wimbledon is 6 feet longer than twice the width. If the court’s perimeter is 228 feet, what are the court’s dimensions?

  10. 28. The length of a rectangular pool is 6 meters less than twice the width. If the pool’s perimeter is 126 meters, what are its dimensions?

  11. 29. The rectangular painting in the figure shown measures 12 inches by 16 inches and is surrounded by a frame of uniform width around the four edges. The perimeter of the rectangle formed by the painting and its frame is 72 inches. Determine the width of the frame.

    The image shows a rectangular painting of length 16 inches and width 12 inches and is surrounded by a frame of uniform width around the four edges labeled x.
  12. 30. The rectangular swimming pool in the figure shown measures 40 feet by 60 feet and is surrounded by a path of uniform width around the four edges. The perimeter of the rectangle formed by the pool and the surrounding path is 248 feet. Determine the width of the path.

    The image shows the rectangular swimming pool measures length of 60 feet and width of 40 feet and is surrounded by a path of uniform width around the four edges labeled x.

  13. 31. The length of a rectangular sign is 3 feet longer than the width. If the sign’s area is 54 square feet, find its length and width.

  14. 32. A rectangular parking lot has a length that is 3 yards greater than the width. The area of the parking lot is 180 square yards. Find the length and the width.

  15. 33. Each side of a square is lengthened by 3 inches. The area of this new, larger square is 64 square inches. Find the length of a side of the original square.

  16. 34. Each side of a square is lengthened by 2 inches. The area of this new, larger square is 36 square inches. Find the length of a side of the original square.

  17. 35. A pool measuring 10 meters by 20 meters is surrounded by a path of uniform width. If the area of the pool and the path combined is 600 square meters, what is the width of the path?

  18. 36. A vacant rectangular lot is being turned into a community vegetable garden measuring 15 meters by 12 meters. A path of uniform width is to surround the garden. If the area of the lot is 378 square meters, find the width of the path surrounding the garden.

  19. 37. As part of a landscaping project, you put in a flower bed measuring 20 feet by 30 feet. To finish off the project, you are putting in a uniform border of pine bark around the outside of the rectangular garden. You have enough pine bark to cover 336 square feet. How wide should the border be?

  20. 38. As part of a landscaping project, you put in a flower bed measuring 10 feet by 12 feet. You plan to surround the bed with a uniform border of low-growing plants that require 1 square foot each when mature. If you have 168 of these plants, how wide a strip around the flower bed should you prepare for the border?

  21. 39. A 20-foot ladder is 15 feet from a house. How far up the house, to the nearest tenth of a foot, does the ladder reach?

  22. 40. The base of a 30-foot ladder is 10 feet from a building. If the ladder reaches the flat roof, how tall, to the nearest tenth of a foot, is the building?

  23. 41. A tree is supported by a wire anchored in the ground 5 feet from its base. The wire is 1 foot longer than the height that it reaches on the tree. Find the length of the wire.

  24. 42. A tree is supported by a wire anchored in the ground 15 feet from its base. The wire is 4 feet longer than the height that it reaches on the tree. Find the length of the wire.

  25. 43. A rectangular piece of land whose length its twice its width has a diagonal distance of 64 yards. How many yards, to the nearest tenth of a yard, does a person save by walking diagonally across the land instead of walking its length and its width?

  26. 44. A rectangular piece of land whose length is three times its width has a diagonal distance of 92 yards. How many yards, to the nearest tenth of a yard, does a person save by walking diagonally across the land instead of walking its length and its width?

  27. 45. A group of people share equally in a $20,000,000 lottery. Before the money is divided, two more winning ticket holders are declared. As a result, each person’s share is reduced by $500,000. How many people were in the original group of winners?

  28. 46. A group of friends agrees to share the cost of a $480,000 vacation condominium equally. Before the purchase is made, four more people join the group and enter the agreement. As a result, each person’s share is reduced by $32,000. How many people were in the original group?

In Exercises 4750, use the formula

Time traveled=Distance traveledAverage velocity.
  1. 47. A car can travel 300 miles in the same amount of time it takes a bus to travel 180 miles. If the average velocity of the bus is 20 miles per hour slower than the average velocity of the car, find the average velocity for each.

  2. 48. A passenger train can travel 240 miles in the same amount of time it takes a freight train to travel 160 miles. If the average velocity of the freight train is 20 miles per hour slower than the average velocity of the passenger train, find the average velocity of each.

  3. 49. You ride your bike to campus a distance of 5 miles and return home on the same route. Going to campus, you ride mostly downhill and average 9 miles per hour faster than on your return trip home. If the round trip takes one hour and ten minutes—that is 76 hours—what is your average velocity on the return trip?

  4. 50. An engine pulls a train 140 miles. Then a second engine, whose average velocity is 5 miles per hour faster than the first engine, takes over and pulls the train 200 miles. The total time required for both engines is 9 hours. Find the average velocity of each engine.

  5. 51. An automobile repair shop charged a customer $1182, listing $357 for parts and the remainder for labor. If the cost of labor is $75 per hour, how many hours of labor did it take to repair the car?

  6. 52. A repair bill on a sailboat came to $2356, including $826 for parts and the remainder for labor. If the cost of labor is $90 per hour, how many hours of labor did it take to repair the sailboat?

  7. 53. For an international telephone call, a telephone company charges $0.43 for the first minute, $0.32 for each additional minute, and a $2.10 service charge. If the cost of a call is $5.73, how long did the person talk?

  8. 54. A job pays an annual salary of $57,900, which includes a holiday bonus of $1500. If paychecks are issued twice a month, what is the gross amount for each paycheck?

  9. 55. You have 35 hits in 140 times at bat. Your batting average is 35140, or 0.25. How many consecutive hits must you get to increase your batting average to 0.30?

  10. 56. You have 30 hits in 120 times at bat. Your batting average is 30120, or 0.25. How many consecutive hits must you get to increase your batting average to 0.28?

Explaining the Concepts

  1. 57. In your own words, describe a step-by-step approach for solving algebraic word problems.

  2. 58. Write an original word problem that can be solved using a linear equation. Then solve the problem.

Critical Thinking Exercises

MAKE SENSE? In Exercises 5961, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 59. By modeling attitudes of college freshmen from 1969 through 2018, I can make precise predictions about the attitudes of the freshman class of 2040.

  2. 60. I find the hardest part in solving a word problem is writing the equation that models the verbal conditions.

  3. 61. After a 35% reduction, a laptop’s price is $780, so I determined the original price, x, by solving x0.35=780.

  4. 62. At the north campus of a performing arts school, 10% of the students are music majors. At the south campus, 90% of the students are music majors. The campuses are merged into one east campus. If 42% of the 1000 students at the east campus are music majors, how many students did the north and south campuses have before the merger?

  5. 63. The price of a dress is reduced by 40%. When the dress still does not sell, it is reduced by 40% of the reduced price. If the price of the dress after both reductions is $72, what was the original price?

  6. 64. If I am three times your age and in 20 years I’ll be twice your age, how old are we?

  7. 65. Suppose that we agree to pay you 8¢ for every problem in this chapter that you solve correctly and fine you 5¢ for every problem done incorrectly. If at the end of 26 problems we do not owe each other any money, how many problems did you solve correctly?

  8. 66. It was wartime when Dick and Jane found out Jane was pregnant. Dick was drafted and made out a will, deciding that $14,000 in a savings account was to be divided between his wife and his child-to-be. Rather strangely, and certainly with gender bias, Dick stipulated that if the child were a boy, he would get twice the amount of the mother’s portion. If it were a girl, the mother would get twice the amount the girl was to receive. We’ll never know what Dick was thinking of, for (as fate would have it) he did not return from war. Jane gave birth to twins—a boy and a girl. How was the money divided?

  9. 67. A thief steals a number of rare plants from a nursery. On the way out, the thief meets three security guards, one after another. To each security guard, the thief is forced to give one-half the plants that he still has, plus two more. Finally, the thief leaves the nursery with one lone palm. How many plants were originally stolen?

Group Exercise

  1. 68. One of the best ways to learn how to solve a word problem in algebra is to design word problems of your own. Creating a word problem makes you very aware of precisely how much information is needed to solve the problem. You must also focus on the best way to present information to a reader and on how much information to give. As you write your problem, you gain skills that will help you solve problems created by others.

    The group should design five different word problems that can be solved using linear equations. All of the problems should be on different topics. For example, the group should not have more than one problem on simple interest. The group should turn in both the problems and their algebraic solutions.

Preview Exercises

Exercises 6971 will help you prepare for the material covered in the next section.

  1. 69. Is 1 a solution of 32x11?

  2. 70. Solve: 2x4=x+5.

  3. 71. Solve: x+34=x23+14.

P.8: Exercise Set

P.8 Exercise Set

Practice and Application Exercises

How will you spend your average life expectancy of 78 years? The bar graph shows the average number of years you will devote to each of your most time-consuming activities. Exercises 12 are based on the data displayed by the graph.

The image shows a bar graph titled, How You Will Spend Your Average Life Expectancy of 78 Years.

Source: U.S. Bureau of Labor Statistics

  1. 1. According to the U.S. Bureau of Labor Statistics, you will devote 37 years to sleeping and watching TV. The number of years sleeping will exceed the number of years watching TV by 19. Over your lifetime, how many years will you spend on each of these activities?

  2. 2. According to the U.S. Bureau of Labor Statistics, you will devote 32 years to sleeping and eating. The number of years sleeping will exceed the number of years eating by 24. Over your lifetime, how many years will you spend on each of these activities?

The bar graph shows median yearly earnings of full-time workers in the United States for people 25 years and over for three occupations and two levels of education. Exercises 34 are based on the data displayed by the graph.

A bar graph titled, Median earning of full- time U.S. workers 25 years and over by occupation and level of education.

Sources: U.S. Census Bureau; Education Pays 2019

  1. 3. The median yearly salary of a general manager with a bachelor’s degree or higher is $31,000 less than twice that of a general manager with just a high school diploma. Combined, two managers with each of these educational attainments earn $149,300. Find the median yearly salary of general managers with each of these levels of education.

  2. 4. The median yearly salary of a retail salesperson with a bachelor’s degree or higher is $14,300 less than twice that of a retail salesperson with just a high school diploma. Combined, two salespeople with each of these educational attainments earn $79,900. Find the median yearly salary of salespeople with each of these levels of education.

Despite booming new car sales with their cha-ching sounds, the average age of vehicles on U.S. roads is not going down. The bar graph shows the average price of new cars in the United States and the average age of cars on U.S. roads for two selected years. Exercises 56 are based on the information displayed by the graph.

A bar graph titled, Average price of new cars and Average age of cars on U.S. roads.

Source: Kelley Blue Book, IHS Automotive/Polk

  1. 5. In 2019, the average price of a new car was $38,900. For the period shown, new-car prices increased by approximately $800 per year. If this trend continues, how many years after 2019 will the price of a new car average $44,500? In which year will this occur?

  2. 6. In 2019, the average age of cars on U.S. roads was 11.8 years. For the period shown, this average age increased by approximately 0.15 year per year. If this trend continues, how many years after 2019 will the average age of vehicles on U.S. roads be 13 years? In which year will this occur?

  3. 7. A new car worth $36,000 is depreciating in value by $4000 per year.

    1. Write a formula that models the car’s value, y, in dollars, after x years.

    2. Use the formula from part (a) to determine after how many years the car’s value will be $12,000.

  4. 8. A new car worth $45,000 is depreciating in value by $5000 per year.

    1. Write a formula that models the car’s value, y, in dollars, after x years.

    2. Use the formula from part (a) to determine after how many years the car’s value will be $10,000.

  5. 9. You are choosing between two gyms. One gym offers membership for a fee of $40 plus a monthly fee of $25. The other offers membership for a fee of $15 plus a monthly fee of $30. After how many months will the total cost at each gym be the same? What will be the total cost for each gym?

  6. 10. Taxi rates are determined by local authorities. In New York City, the night-time cost of a taxi includes a base fee of $3 plus a charge of $1.56 per kilometer. In Boston, regardless of the time of day, the base fee is $2.60 with a charge of $1.75 per kilometer. For how many kilometers will the cost of a night-time taxi ride in each city be the same? Round to the nearest kilometer. What will the cost be in each city for the rounded number of kilometers?

  7. 11. A transponder for a toll bridge costs $27.50. With the transponder, the toll is $5 each time you cross the bridge. The only other option is toll-by-plate, for which the toll is $6.25 each time you cross the bridge with an additional administrative fee of $1.25 for each crossing. How many times would you need to cross the bridge for the costs of the two toll options to be the same?

  8. 12. An electronic pass for a toll road costs $30. The toll is normally $5.00 but is reduced by 30% for people who have purchased the electronic pass. Determine the number of times the road must be used so that the total cost without the pass is the same as the total cost with the pass.

  9. 13. In 2010, there were 13,300 students at college A, with a projected enrollment increase of 1000 students per year. In the same year, there were 26,800 students at college B, with a projected enrollment decline of 500 students per year. According to these projections, when did the colleges have the same enrollment? What was the enrollment in each college at that time?

  10. 14. In 2000, the population of Greece was 10,600,000, with projections of a population decrease of 28,000 people per year. In the same year, the population of Belgium was 10,200,000, with projections of a population decrease of 12,000 people per year. (Source: United Nations) According to these projections, when will the two countries have the same population? What will be the population at that time?

  11. 15. After a 20% reduction, you purchase a television for $336. What was the television’s price before the reduction?

  12. 16. After a 30% reduction, you purchase wireless earbuds for $90.30. What was the earbuds’ price before the reduction?

  13. 17. Including a 10.5% hotel tax, your room in San Diego cost $216.58 per night. Find the nightly cost before the tax was added.

  14. 18. Including a 17.4% hotel tax, your room in Chicago cost $287.63 per night. Find the nightly cost before the tax was added.

Exercises 1920 involve markup, the amount added to the dealer’s cost of an item to arrive at the selling price of that item.

  1. 19. The selling price of a refrigerator is $1198. If the markup is 25% of the dealer’s cost, what is the dealer’s cost of the refrigerator?

  2. 20. The selling price of a scientific calculator is $15. If the markup is 25% of the dealer’s cost, what is the dealer’s cost of the calculator?

  3. 21. You invested $20,000 in two accounts paying 1.45% and 1.59% annual interest. If the total interest earned for the year was $307.50, how much was invested at each rate?

  4. 22. You invested $30,000 in two accounts paying 2.19% and 2.45% annual interest. If the total interest earned for the year was $705.88, how much was invested at each rate?

  5. 23. Things did not go quite as planned. You invested $10,000, part of it in a stock that realized a 12% gain. However, the rest of the money suffered a 5% loss. If you had an overall gain of $520, how much was invested at each rate?

  6. 24. Things did not go quite as planned. You invested $15,000, part of it in a stock that realized a 15% gain. However, the rest of the money suffered a 7% loss. If you had an overall gain of $1590, how much was invested at each rate?

  7. 25. A rectangular soccer field is twice as long as it is wide. If the perimeter of the soccer field is 300 yards, what are its dimensions?

  8. 26. A rectangular swimming pool is three times as long as it is wide. If the perimeter of the pool is 320 feet, what are its dimensions?

  9. 27. The length of the rectangular tennis court at Wimbledon is 6 feet longer than twice the width. If the court’s perimeter is 228 feet, what are the court’s dimensions?

  10. 28. The length of a rectangular pool is 6 meters less than twice the width. If the pool’s perimeter is 126 meters, what are its dimensions?

  11. 29. The rectangular painting in the figure shown measures 12 inches by 16 inches and is surrounded by a frame of uniform width around the four edges. The perimeter of the rectangle formed by the painting and its frame is 72 inches. Determine the width of the frame.

    The image shows a rectangular painting of length 16 inches and width 12 inches and is surrounded by a frame of uniform width around the four edges labeled x.
  12. 30. The rectangular swimming pool in the figure shown measures 40 feet by 60 feet and is surrounded by a path of uniform width around the four edges. The perimeter of the rectangle formed by the pool and the surrounding path is 248 feet. Determine the width of the path.

    The image shows the rectangular swimming pool measures length of 60 feet and width of 40 feet and is surrounded by a path of uniform width around the four edges labeled x.

  13. 31. The length of a rectangular sign is 3 feet longer than the width. If the sign’s area is 54 square feet, find its length and width.

  14. 32. A rectangular parking lot has a length that is 3 yards greater than the width. The area of the parking lot is 180 square yards. Find the length and the width.

  15. 33. Each side of a square is lengthened by 3 inches. The area of this new, larger square is 64 square inches. Find the length of a side of the original square.

  16. 34. Each side of a square is lengthened by 2 inches. The area of this new, larger square is 36 square inches. Find the length of a side of the original square.

  17. 35. A pool measuring 10 meters by 20 meters is surrounded by a path of uniform width. If the area of the pool and the path combined is 600 square meters, what is the width of the path?

  18. 36. A vacant rectangular lot is being turned into a community vegetable garden measuring 15 meters by 12 meters. A path of uniform width is to surround the garden. If the area of the lot is 378 square meters, find the width of the path surrounding the garden.

  19. 37. As part of a landscaping project, you put in a flower bed measuring 20 feet by 30 feet. To finish off the project, you are putting in a uniform border of pine bark around the outside of the rectangular garden. You have enough pine bark to cover 336 square feet. How wide should the border be?

  20. 38. As part of a landscaping project, you put in a flower bed measuring 10 feet by 12 feet. You plan to surround the bed with a uniform border of low-growing plants that require 1 square foot each when mature. If you have 168 of these plants, how wide a strip around the flower bed should you prepare for the border?

  21. 39. A 20-foot ladder is 15 feet from a house. How far up the house, to the nearest tenth of a foot, does the ladder reach?

  22. 40. The base of a 30-foot ladder is 10 feet from a building. If the ladder reaches the flat roof, how tall, to the nearest tenth of a foot, is the building?

  23. 41. A tree is supported by a wire anchored in the ground 5 feet from its base. The wire is 1 foot longer than the height that it reaches on the tree. Find the length of the wire.

  24. 42. A tree is supported by a wire anchored in the ground 15 feet from its base. The wire is 4 feet longer than the height that it reaches on the tree. Find the length of the wire.

  25. 43. A rectangular piece of land whose length its twice its width has a diagonal distance of 64 yards. How many yards, to the nearest tenth of a yard, does a person save by walking diagonally across the land instead of walking its length and its width?

  26. 44. A rectangular piece of land whose length is three times its width has a diagonal distance of 92 yards. How many yards, to the nearest tenth of a yard, does a person save by walking diagonally across the land instead of walking its length and its width?

  27. 45. A group of people share equally in a $20,000,000 lottery. Before the money is divided, two more winning ticket holders are declared. As a result, each person’s share is reduced by $500,000. How many people were in the original group of winners?

  28. 46. A group of friends agrees to share the cost of a $480,000 vacation condominium equally. Before the purchase is made, four more people join the group and enter the agreement. As a result, each person’s share is reduced by $32,000. How many people were in the original group?

In Exercises 4750, use the formula

Time traveled=Distance traveledAverage velocity.
  1. 47. A car can travel 300 miles in the same amount of time it takes a bus to travel 180 miles. If the average velocity of the bus is 20 miles per hour slower than the average velocity of the car, find the average velocity for each.

  2. 48. A passenger train can travel 240 miles in the same amount of time it takes a freight train to travel 160 miles. If the average velocity of the freight train is 20 miles per hour slower than the average velocity of the passenger train, find the average velocity of each.

  3. 49. You ride your bike to campus a distance of 5 miles and return home on the same route. Going to campus, you ride mostly downhill and average 9 miles per hour faster than on your return trip home. If the round trip takes one hour and ten minutes—that is 76 hours—what is your average velocity on the return trip?

  4. 50. An engine pulls a train 140 miles. Then a second engine, whose average velocity is 5 miles per hour faster than the first engine, takes over and pulls the train 200 miles. The total time required for both engines is 9 hours. Find the average velocity of each engine.

  5. 51. An automobile repair shop charged a customer $1182, listing $357 for parts and the remainder for labor. If the cost of labor is $75 per hour, how many hours of labor did it take to repair the car?

  6. 52. A repair bill on a sailboat came to $2356, including $826 for parts and the remainder for labor. If the cost of labor is $90 per hour, how many hours of labor did it take to repair the sailboat?

  7. 53. For an international telephone call, a telephone company charges $0.43 for the first minute, $0.32 for each additional minute, and a $2.10 service charge. If the cost of a call is $5.73, how long did the person talk?

  8. 54. A job pays an annual salary of $57,900, which includes a holiday bonus of $1500. If paychecks are issued twice a month, what is the gross amount for each paycheck?

  9. 55. You have 35 hits in 140 times at bat. Your batting average is 35140, or 0.25. How many consecutive hits must you get to increase your batting average to 0.30?

  10. 56. You have 30 hits in 120 times at bat. Your batting average is 30120, or 0.25. How many consecutive hits must you get to increase your batting average to 0.28?

Explaining the Concepts

  1. 57. In your own words, describe a step-by-step approach for solving algebraic word problems.

  2. 58. Write an original word problem that can be solved using a linear equation. Then solve the problem.

Critical Thinking Exercises

MAKE SENSE? In Exercises 5961, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 59. By modeling attitudes of college freshmen from 1969 through 2018, I can make precise predictions about the attitudes of the freshman class of 2040.

  2. 60. I find the hardest part in solving a word problem is writing the equation that models the verbal conditions.

  3. 61. After a 35% reduction, a laptop’s price is $780, so I determined the original price, x, by solving x0.35=780.

  4. 62. At the north campus of a performing arts school, 10% of the students are music majors. At the south campus, 90% of the students are music majors. The campuses are merged into one east campus. If 42% of the 1000 students at the east campus are music majors, how many students did the north and south campuses have before the merger?

  5. 63. The price of a dress is reduced by 40%. When the dress still does not sell, it is reduced by 40% of the reduced price. If the price of the dress after both reductions is $72, what was the original price?

  6. 64. If I am three times your age and in 20 years I’ll be twice your age, how old are we?

  7. 65. Suppose that we agree to pay you 8¢ for every problem in this chapter that you solve correctly and fine you 5¢ for every problem done incorrectly. If at the end of 26 problems we do not owe each other any money, how many problems did you solve correctly?

  8. 66. It was wartime when Dick and Jane found out Jane was pregnant. Dick was drafted and made out a will, deciding that $14,000 in a savings account was to be divided between his wife and his child-to-be. Rather strangely, and certainly with gender bias, Dick stipulated that if the child were a boy, he would get twice the amount of the mother’s portion. If it were a girl, the mother would get twice the amount the girl was to receive. We’ll never know what Dick was thinking of, for (as fate would have it) he did not return from war. Jane gave birth to twins—a boy and a girl. How was the money divided?

  9. 67. A thief steals a number of rare plants from a nursery. On the way out, the thief meets three security guards, one after another. To each security guard, the thief is forced to give one-half the plants that he still has, plus two more. Finally, the thief leaves the nursery with one lone palm. How many plants were originally stolen?

Group Exercise

  1. 68. One of the best ways to learn how to solve a word problem in algebra is to design word problems of your own. Creating a word problem makes you very aware of precisely how much information is needed to solve the problem. You must also focus on the best way to present information to a reader and on how much information to give. As you write your problem, you gain skills that will help you solve problems created by others.

    The group should design five different word problems that can be solved using linear equations. All of the problems should be on different topics. For example, the group should not have more than one problem on simple interest. The group should turn in both the problems and their algebraic solutions.

Preview Exercises

Exercises 6971 will help you prepare for the material covered in the next section.

  1. 69. Is 1 a solution of 32x11?

  2. 70. Solve: 2x4=x+5.

  3. 71. Solve: x+34=x23+14.

Section P.9: Linear Inequalities and Absolute Value Inequalities

Section P.9 Linear Inequalities and Absolute Value Inequalities

Learning Objectives

What You’ll Learn

  1. 1 Use interval notation.

  2. 2 Find intersections and unions of intervals.

  3. 3 Solve linear inequalities.

  4. 4 Solve compound inequalities.

  5. 5 Solve absolute value inequalities.

Rent-a-Heap, a car rental company, charges $125 per week plus $0.20 per mile to rent one of their cars. Suppose you are limited by how much money you can spend for the week: You can spend at most $335. If we let x represent the number of miles you drive the heap in a week, we can write an inequality that models the given conditions.

The image shows a mathematical expression 125 + 0.20 x is less than or equal to 335.
P.10-207 Full Alternative Text

Placing an inequality symbol between a polynomial of degree 1 and a constant results in a linear inequality in one variable. In this section, we will study how to solve linear inequalities such as 125+0.20x335. Solving an inequality is the process of finding the set of numbers that make the inequality a true statement. These numbers are called the solutions of the inequality and we say that they satisfy the inequality. The set of all solutions is called the solution set of the inequality. Set-builder notation and a new notation, called interval notation, are used to represent these solution sets. We begin this section by looking at interval notation.

Objective 1: Use interval notation

Interval Notation

  1. Objective 1Use interval notation.

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Some sets of real numbers can be represented using interval notation. Suppose that a and b are two real numbers such that a<b.

A chart shows the interval notation and respective graph.
P.10-208 Full Alternative Text

Parentheses and Brackets in Interval Notation

Parentheses indicate endpoints that are not included in an interval. Square brackets indicate endpoints that are included in an interval. Parentheses are always used with ∞ or .

Table P.7 lists nine possible types of intervals used to describe subsets of real numbers.

Table P.7 Intervals on the Real Number Line

A table shows intervals on the real number line.
Table P.7 Full Alternative Text

Example 1 Using Interval Notation

Express each interval in set-builder notation and graph:

  1. (1, 4]

  2. [2.5, 4]

  3. (4, ).

Solution

  1. (1, 4]={x|1<x4}

    A number line ranging from negative 4 to 4 is shaded from left parenthesis at negative 1 to right bracket at 4.
  2. [2.5, 4]={x|2.5x4}

    A number line ranging from negative 4 to 4 is shaded from the left bracket at 2.5 to the right bracket at 4.
  3. (4, )={x|x>4}

    A number line ranging from negative 4 to 4 is shaded right to the left parenthesis at negative 4.

Check Point 1

  • Express each interval in set-builder notation and graph:

    1. [2, 5)

    2. [1, 3.5]

    3. (, 1).

Objective 2: Find intersections and unions of intervals

Intersections and Unions of Intervals

  1. Objective 2Find intersections and unions of intervals.

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In Section P.1, we learned how to find intersections and unions of sets. Recall that AB (A intersection B) is the set of elements common to both set A and set B. By contrast, AB (A union B) is the set of elements in set A or in set B or in both sets.

Because intervals represent sets, it is possible to find their intersections and unions. Graphs are helpful in this process.

Finding Intersections and Unions of Two Intervals

  1. Graph each interval on a number line.

    1. To find the intersection, take the portion of the number line that the two graphs have in common.

    2. To find the union, take the portion of the number line representing the total collection of numbers in the two graphs.

Example 2 Finding Intersections and Unions of Intervals

Use graphs to find each set:

  1. (1, 4)[2, 8]

  2. (1, 4)[2, 8].

Solution

  1. (1, 4)[2, 8], the intersection of the intervals (1, 4) and [2, 8], consists of the numbers that are in both intervals.

    Two number lines.

    To find (1, 4)[2, 8], take the portion of the number line that the two graphs have in common.

    The image shows a number line.

    Thus, (1, 4)[2, 8]=[2, 4).

  2. (1, 4)[2, 8], the union of the intervals (1, 4) and [2, 8], consists of the numbers that are in either one interval or the other (or both).

    The image shows a number line.

    To find (1, 4)[2, 8], take the portion of the number line representing the total collection of numbers in the two graphs.

    The image shows a number line.

    Thus, (1, 4)[2, 8]=(1, 8].

Check Point 2

  • Use graphs to find each set:

    1. [1, 3](2, 6)

    2. [1, 3](2, 6).

Objective 3: Solve linear inequalities

Solving Linear Inequalities in One Variable

  1. Objective 3Solve linear inequalities.

Watch Video

We know that a linear equation in x can be expressed as ax+b=0. A linear inequality in x can be written in one of the following forms:

ax+b<0, ax+b0, ax+b>0, ax+b0.

In each form, a0.

Back to our question that opened this section: How many miles can you drive your Rent-a-Heap car if you can spend at most $335? We answer the question by solving

0.20x+125335

for x. The solution procedure is nearly identical to that for solving

0.20x+125=335.

Our goal is to get x by itself on the left side. We do this by subtracting 125 from both sides to isolate 0.20x:

0.20x+125335This is the given inequality.0.20x+125125335125Subtract 125 from both sides.0.20x210.Simplify.

Finally, we isolate x from 0.20x by dividing both sides of the inequality by 0.20:

0.20x0.202100.20Divide both sides by 0.20.x1050.Simplify.

With at most $335 to spend, you can travel at most 1050 miles.

We started with the inequality 0.20x+125335 and obtained the inequality x1050 in the final step. These inequalities have the same solution set, namely, {x|x1050}. Inequalities such as these, with the same solution set, are said to be equivalent.

We isolated x from 0.20x by dividing both sides of 0.20x210 by 0.20, a positive number. Let’s see what happens if we divide both sides of an inequality by a negative number. Consider the inequality 10<14. Divide 10 and 14 by 2:

102=5and142=7.

Because 5 lies to the right of 7 on the number line, 5 is greater than 7:

5>7.

Notice that the direction of the inequality symbol is reversed:

A two way arrow between 10 is less than 14 and negative 5 is greater than negative 7. A note reads, dividing by negative 2 changes the direction of the inequality symbol.

In general, when we multiply or divide both sides of an inequality by a negative number, the direction of the inequality symbol is reversed. When we reverse the direction of the inequality symbol, we say that we change the sense of the inequality.

We can isolate a variable in a linear inequality in the same way we isolate a variable in a linear equation. The properties on the next page are used to create equivalent inequalities.

Properties of Inequalities

Property The Property in Words Example

The Addition Property of Inequality

If a<b, then a+c<b+c.

If a<b, then ac<bc.

If the same quantity is added to or subtracted from both sides of an inequality, the resulting inequality is equivalent to the original one.

2x+3<7

Subtract 3:

2x+33<73.

Simplify:

2x<4.

The Positive Multiplication Property of Inequality

If a<b and c is positive, then ac<bc.

If a<b and c is positive, then ac<bc.

If we multiply or divide both sides of an inequality by the same positive quantity, the resulting inequality is equivalent to the original one.

2x<4

Divide by 2:

2x2<42.

Simplify:

x<2.

The Negative Multiplication Property of Inequality

If a<b and c is negative, then ac>bc.

If a<b and c is negative, then ac>bc.

If we multiply or divide both sides of an inequality by the same negative quantity and reverse the direction of the inequality symbol, the resulting inequality is equivalent to the original one.

4x<20

Divide by 4 and change the sense of the inequality:

4x4>204.

Simplify:

x>5.

Example 3 Solving a Linear Inequality

Solve and graph the solution set on a number line:

32x11.

Solution

32x11This is the given inequality.32x3113Subtract 3 from both sides.2x8Simplify.2x282Divide both sides by 2  andchange the sense of the inequality.x4Simplify.

The solution set of 32x11, or equivalently x4, consists of all real numbers that are greater than or equal to 4, expressed as {x|x4} in set-builder notation. The interval notation for this solution set is [4, ). The graph of the solution set is shown as follows:

A number line ranging from negative 5 to 5 is shaded right to a left bracket at 4.

Check Point 3

  • Solve and graph the solution set on a number line:

    23x5.

Example 4 Solving a Linear Inequality

Solve and graph the solution set on a number line:

2x4>x+5.

Solution

  1. Step 1 SIMPLIFY EACH SIDE. Because each side is already simplified, we can skip this step.

  2. Step 2 COLLECT VARIABLE TERMS ON ONE SIDE AND CONSTANT TERMS ON THE OTHER SIDE. We will collect variable terms of 2x4>x+5 on the left and constant terms on the right.

    2x4>x+5This  is  the  given  inequlity.2x4x>x+5xSubtract  x  from  both  sides.3x4>5Simplify.3x4+4>5+4Add  4  to  both  sides.3x>9Simplify.
  3. Step 3 ISOLATE THE VARIABLE AND SOLVE. We isolate the variable, x, by dividing both sides by 3. Because we are dividing by a negative number, we must reverse the inequality symbol.

    3x3<93Divide both sides by 3 and   change the sense of the inequality.x<3   Simplify.
  4. Step 4 EXPRESS THE SOLUTION SET IN SET-BUILDER OR INTERVAL NOTATION AND GRAPH THE SET ON A NUMBER LINE. The solution set consists of all real numbers that are less than 3, expressed in set-builder notation as {x|x<3}. The interval notation for this solution set is (, 3). The graph of the solution set is shown as follows:

    A number line ranging from negative 5 to 5 is shaded left to a right parenthesis at negative 3.

Check Point 4

  • Solve and graph the solution set on a number line: 3x+1>7x15.

If an inequality contains fractions with constants in the denominators, begin by multiplying both sides by the least common denominator. This will clear the inequality of fractions.

Example 5 Solving a Linear Inequality Containing Fractions

Solve and graph the solution set on a number line:

x+34x23+14.

Solution

The denominators are 4, 3, and 4. The least common denominator is 12. We begin by multiplying both sides of the inequality by 12.

The image shows the steps to solve an inequality.

Now that the fractions have been cleared, we follow the four steps that we used in the previous example.

  1. Step 1 SIMPLIFY EACH SIDE.

    3(x+3)4(x2)+3This is the inequality with the fractions cleared.3x+94x8+3Use the distributive property.3x+94x5Simplify.
  2. Step 2 COLLECT VARIABLE TERMS ON ONE SIDE AND CONSTANT TERMS ON THE OTHER SIDE. We will collect variable terms of 3x+94x5 on the left and constant terms on the right.

    3x+94x4x54xSubtract 4x from both sides.x+95Simplify.x+9959Subtract 9 from both sides.x14Simplify.
  3. Step 3 ISOLATE THE VARIABLE AND SOLVE. To isolate x, we must eliminate the negative sign in front of the x. Because x means 1x, we can do this by multiplying (or dividing) both sides of the inequality by 1. We are multiplying by a negative number. Thus, we must reverse the direction of the inequality symbol.

    (1)(x)(1)(14) Multiply both sides by 1 and change the sense of the inequality.x14Simplify.
  4. Step 4 EXPRESS THE SOLUTION SET IN SET-BUILDER OR INTERVAL NOTATION AND GRAPH THE SET ON A NUMBER LINE. The solution set consists of all real numbers that are less than or equal to 14, expressed in set-builder notation as {x|x14}. The interval notation for this solution set is (, 14]. The graph of the solution set is shown as follows:

    A number line ranging from 5 to 15 is shaded left to a right bracket at 14.

Check Point 5

  • Solve and graph the solution set on a number line:

    x42x23+56.
Objective 3: Solve linear inequalities

Solving Linear Inequalities in One Variable

  1. Objective 3Solve linear inequalities.

Watch Video

We know that a linear equation in x can be expressed as ax+b=0. A linear inequality in x can be written in one of the following forms:

ax+b<0, ax+b0, ax+b>0, ax+b0.

In each form, a0.

Back to our question that opened this section: How many miles can you drive your Rent-a-Heap car if you can spend at most $335? We answer the question by solving

0.20x+125335

for x. The solution procedure is nearly identical to that for solving

0.20x+125=335.

Our goal is to get x by itself on the left side. We do this by subtracting 125 from both sides to isolate 0.20x:

0.20x+125335This is the given inequality.0.20x+125125335125Subtract 125 from both sides.0.20x210.Simplify.

Finally, we isolate x from 0.20x by dividing both sides of the inequality by 0.20:

0.20x0.202100.20Divide both sides by 0.20.x1050.Simplify.

With at most $335 to spend, you can travel at most 1050 miles.

We started with the inequality 0.20x+125335 and obtained the inequality x1050 in the final step. These inequalities have the same solution set, namely, {x|x1050}. Inequalities such as these, with the same solution set, are said to be equivalent.

We isolated x from 0.20x by dividing both sides of 0.20x210 by 0.20, a positive number. Let’s see what happens if we divide both sides of an inequality by a negative number. Consider the inequality 10<14. Divide 10 and 14 by 2:

102=5and142=7.

Because 5 lies to the right of 7 on the number line, 5 is greater than 7:

5>7.

Notice that the direction of the inequality symbol is reversed:

A two way arrow between 10 is less than 14 and negative 5 is greater than negative 7. A note reads, dividing by negative 2 changes the direction of the inequality symbol.

In general, when we multiply or divide both sides of an inequality by a negative number, the direction of the inequality symbol is reversed. When we reverse the direction of the inequality symbol, we say that we change the sense of the inequality.

We can isolate a variable in a linear inequality in the same way we isolate a variable in a linear equation. The properties on the next page are used to create equivalent inequalities.

Properties of Inequalities

Property The Property in Words Example

The Addition Property of Inequality

If a<b, then a+c<b+c.

If a<b, then ac<bc.

If the same quantity is added to or subtracted from both sides of an inequality, the resulting inequality is equivalent to the original one.

2x+3<7

Subtract 3:

2x+33<73.

Simplify:

2x<4.

The Positive Multiplication Property of Inequality

If a<b and c is positive, then ac<bc.

If a<b and c is positive, then ac<bc.

If we multiply or divide both sides of an inequality by the same positive quantity, the resulting inequality is equivalent to the original one.

2x<4

Divide by 2:

2x2<42.

Simplify:

x<2.

The Negative Multiplication Property of Inequality

If a<b and c is negative, then ac>bc.

If a<b and c is negative, then ac>bc.

If we multiply or divide both sides of an inequality by the same negative quantity and reverse the direction of the inequality symbol, the resulting inequality is equivalent to the original one.

4x<20

Divide by 4 and change the sense of the inequality:

4x4>204.

Simplify:

x>5.

Example 3 Solving a Linear Inequality

Solve and graph the solution set on a number line:

32x11.

Solution

32x11This is the given inequality.32x3113Subtract 3 from both sides.2x8Simplify.2x282Divide both sides by 2  andchange the sense of the inequality.x4Simplify.

The solution set of 32x11, or equivalently x4, consists of all real numbers that are greater than or equal to 4, expressed as {x|x4} in set-builder notation. The interval notation for this solution set is [4, ). The graph of the solution set is shown as follows:

A number line ranging from negative 5 to 5 is shaded right to a left bracket at 4.

Check Point 3

  • Solve and graph the solution set on a number line:

    23x5.

Example 4 Solving a Linear Inequality

Solve and graph the solution set on a number line:

2x4>x+5.

Solution

  1. Step 1 SIMPLIFY EACH SIDE. Because each side is already simplified, we can skip this step.

  2. Step 2 COLLECT VARIABLE TERMS ON ONE SIDE AND CONSTANT TERMS ON THE OTHER SIDE. We will collect variable terms of 2x4>x+5 on the left and constant terms on the right.

    2x4>x+5This  is  the  given  inequlity.2x4x>x+5xSubtract  x  from  both  sides.3x4>5Simplify.3x4+4>5+4Add  4  to  both  sides.3x>9Simplify.
  3. Step 3 ISOLATE THE VARIABLE AND SOLVE. We isolate the variable, x, by dividing both sides by 3. Because we are dividing by a negative number, we must reverse the inequality symbol.

    3x3<93Divide both sides by 3 and   change the sense of the inequality.x<3   Simplify.
  4. Step 4 EXPRESS THE SOLUTION SET IN SET-BUILDER OR INTERVAL NOTATION AND GRAPH THE SET ON A NUMBER LINE. The solution set consists of all real numbers that are less than 3, expressed in set-builder notation as {x|x<3}. The interval notation for this solution set is (, 3). The graph of the solution set is shown as follows:

    A number line ranging from negative 5 to 5 is shaded left to a right parenthesis at negative 3.

Check Point 4

  • Solve and graph the solution set on a number line: 3x+1>7x15.

If an inequality contains fractions with constants in the denominators, begin by multiplying both sides by the least common denominator. This will clear the inequality of fractions.

Example 5 Solving a Linear Inequality Containing Fractions

Solve and graph the solution set on a number line:

x+34x23+14.

Solution

The denominators are 4, 3, and 4. The least common denominator is 12. We begin by multiplying both sides of the inequality by 12.

The image shows the steps to solve an inequality.

Now that the fractions have been cleared, we follow the four steps that we used in the previous example.

  1. Step 1 SIMPLIFY EACH SIDE.

    3(x+3)4(x2)+3This is the inequality with the fractions cleared.3x+94x8+3Use the distributive property.3x+94x5Simplify.
  2. Step 2 COLLECT VARIABLE TERMS ON ONE SIDE AND CONSTANT TERMS ON THE OTHER SIDE. We will collect variable terms of 3x+94x5 on the left and constant terms on the right.

    3x+94x4x54xSubtract 4x from both sides.x+95Simplify.x+9959Subtract 9 from both sides.x14Simplify.
  3. Step 3 ISOLATE THE VARIABLE AND SOLVE. To isolate x, we must eliminate the negative sign in front of the x. Because x means 1x, we can do this by multiplying (or dividing) both sides of the inequality by 1. We are multiplying by a negative number. Thus, we must reverse the direction of the inequality symbol.

    (1)(x)(1)(14) Multiply both sides by 1 and change the sense of the inequality.x14Simplify.
  4. Step 4 EXPRESS THE SOLUTION SET IN SET-BUILDER OR INTERVAL NOTATION AND GRAPH THE SET ON A NUMBER LINE. The solution set consists of all real numbers that are less than or equal to 14, expressed in set-builder notation as {x|x14}. The interval notation for this solution set is (, 14]. The graph of the solution set is shown as follows:

    A number line ranging from 5 to 15 is shaded left to a right bracket at 14.

Check Point 5

  • Solve and graph the solution set on a number line:

    x42x23+56.
Objective 3: Solve linear inequalities

Solving Linear Inequalities in One Variable

  1. Objective 3Solve linear inequalities.

Watch Video

We know that a linear equation in x can be expressed as ax+b=0. A linear inequality in x can be written in one of the following forms:

ax+b<0, ax+b0, ax+b>0, ax+b0.

In each form, a0.

Back to our question that opened this section: How many miles can you drive your Rent-a-Heap car if you can spend at most $335? We answer the question by solving

0.20x+125335

for x. The solution procedure is nearly identical to that for solving

0.20x+125=335.

Our goal is to get x by itself on the left side. We do this by subtracting 125 from both sides to isolate 0.20x:

0.20x+125335This is the given inequality.0.20x+125125335125Subtract 125 from both sides.0.20x210.Simplify.

Finally, we isolate x from 0.20x by dividing both sides of the inequality by 0.20:

0.20x0.202100.20Divide both sides by 0.20.x1050.Simplify.

With at most $335 to spend, you can travel at most 1050 miles.

We started with the inequality 0.20x+125335 and obtained the inequality x1050 in the final step. These inequalities have the same solution set, namely, {x|x1050}. Inequalities such as these, with the same solution set, are said to be equivalent.

We isolated x from 0.20x by dividing both sides of 0.20x210 by 0.20, a positive number. Let’s see what happens if we divide both sides of an inequality by a negative number. Consider the inequality 10<14. Divide 10 and 14 by 2:

102=5and142=7.

Because 5 lies to the right of 7 on the number line, 5 is greater than 7:

5>7.

Notice that the direction of the inequality symbol is reversed:

A two way arrow between 10 is less than 14 and negative 5 is greater than negative 7. A note reads, dividing by negative 2 changes the direction of the inequality symbol.

In general, when we multiply or divide both sides of an inequality by a negative number, the direction of the inequality symbol is reversed. When we reverse the direction of the inequality symbol, we say that we change the sense of the inequality.

We can isolate a variable in a linear inequality in the same way we isolate a variable in a linear equation. The properties on the next page are used to create equivalent inequalities.

Properties of Inequalities

Property The Property in Words Example

The Addition Property of Inequality

If a<b, then a+c<b+c.

If a<b, then ac<bc.

If the same quantity is added to or subtracted from both sides of an inequality, the resulting inequality is equivalent to the original one.

2x+3<7

Subtract 3:

2x+33<73.

Simplify:

2x<4.

The Positive Multiplication Property of Inequality

If a<b and c is positive, then ac<bc.

If a<b and c is positive, then ac<bc.

If we multiply or divide both sides of an inequality by the same positive quantity, the resulting inequality is equivalent to the original one.

2x<4

Divide by 2:

2x2<42.

Simplify:

x<2.

The Negative Multiplication Property of Inequality

If a<b and c is negative, then ac>bc.

If a<b and c is negative, then ac>bc.

If we multiply or divide both sides of an inequality by the same negative quantity and reverse the direction of the inequality symbol, the resulting inequality is equivalent to the original one.

4x<20

Divide by 4 and change the sense of the inequality:

4x4>204.

Simplify:

x>5.

Example 3 Solving a Linear Inequality

Solve and graph the solution set on a number line:

32x11.

Solution

32x11This is the given inequality.32x3113Subtract 3 from both sides.2x8Simplify.2x282Divide both sides by 2  andchange the sense of the inequality.x4Simplify.

The solution set of 32x11, or equivalently x4, consists of all real numbers that are greater than or equal to 4, expressed as {x|x4} in set-builder notation. The interval notation for this solution set is [4, ). The graph of the solution set is shown as follows:

A number line ranging from negative 5 to 5 is shaded right to a left bracket at 4.

Check Point 3

  • Solve and graph the solution set on a number line:

    23x5.

Example 4 Solving a Linear Inequality

Solve and graph the solution set on a number line:

2x4>x+5.

Solution

  1. Step 1 SIMPLIFY EACH SIDE. Because each side is already simplified, we can skip this step.

  2. Step 2 COLLECT VARIABLE TERMS ON ONE SIDE AND CONSTANT TERMS ON THE OTHER SIDE. We will collect variable terms of 2x4>x+5 on the left and constant terms on the right.

    2x4>x+5This  is  the  given  inequlity.2x4x>x+5xSubtract  x  from  both  sides.3x4>5Simplify.3x4+4>5+4Add  4  to  both  sides.3x>9Simplify.
  3. Step 3 ISOLATE THE VARIABLE AND SOLVE. We isolate the variable, x, by dividing both sides by 3. Because we are dividing by a negative number, we must reverse the inequality symbol.

    3x3<93Divide both sides by 3 and   change the sense of the inequality.x<3   Simplify.
  4. Step 4 EXPRESS THE SOLUTION SET IN SET-BUILDER OR INTERVAL NOTATION AND GRAPH THE SET ON A NUMBER LINE. The solution set consists of all real numbers that are less than 3, expressed in set-builder notation as {x|x<3}. The interval notation for this solution set is (, 3). The graph of the solution set is shown as follows:

    A number line ranging from negative 5 to 5 is shaded left to a right parenthesis at negative 3.

Check Point 4

  • Solve and graph the solution set on a number line: 3x+1>7x15.

If an inequality contains fractions with constants in the denominators, begin by multiplying both sides by the least common denominator. This will clear the inequality of fractions.

Example 5 Solving a Linear Inequality Containing Fractions

Solve and graph the solution set on a number line:

x+34x23+14.

Solution

The denominators are 4, 3, and 4. The least common denominator is 12. We begin by multiplying both sides of the inequality by 12.

The image shows the steps to solve an inequality.

Now that the fractions have been cleared, we follow the four steps that we used in the previous example.

  1. Step 1 SIMPLIFY EACH SIDE.

    3(x+3)4(x2)+3This is the inequality with the fractions cleared.3x+94x8+3Use the distributive property.3x+94x5Simplify.
  2. Step 2 COLLECT VARIABLE TERMS ON ONE SIDE AND CONSTANT TERMS ON THE OTHER SIDE. We will collect variable terms of 3x+94x5 on the left and constant terms on the right.

    3x+94x4x54xSubtract 4x from both sides.x+95Simplify.x+9959Subtract 9 from both sides.x14Simplify.
  3. Step 3 ISOLATE THE VARIABLE AND SOLVE. To isolate x, we must eliminate the negative sign in front of the x. Because x means 1x, we can do this by multiplying (or dividing) both sides of the inequality by 1. We are multiplying by a negative number. Thus, we must reverse the direction of the inequality symbol.

    (1)(x)(1)(14) Multiply both sides by 1 and change the sense of the inequality.x14Simplify.
  4. Step 4 EXPRESS THE SOLUTION SET IN SET-BUILDER OR INTERVAL NOTATION AND GRAPH THE SET ON A NUMBER LINE. The solution set consists of all real numbers that are less than or equal to 14, expressed in set-builder notation as {x|x14}. The interval notation for this solution set is (, 14]. The graph of the solution set is shown as follows:

    A number line ranging from 5 to 15 is shaded left to a right bracket at 14.

Check Point 5

  • Solve and graph the solution set on a number line:

    x42x23+56.
Objective 3: Solve linear inequalities

Solving Linear Inequalities in One Variable

  1. Objective 3Solve linear inequalities.

Watch Video

We know that a linear equation in x can be expressed as ax+b=0. A linear inequality in x can be written in one of the following forms:

ax+b<0, ax+b0, ax+b>0, ax+b0.

In each form, a0.

Back to our question that opened this section: How many miles can you drive your Rent-a-Heap car if you can spend at most $335? We answer the question by solving

0.20x+125335

for x. The solution procedure is nearly identical to that for solving

0.20x+125=335.

Our goal is to get x by itself on the left side. We do this by subtracting 125 from both sides to isolate 0.20x:

0.20x+125335This is the given inequality.0.20x+125125335125Subtract 125 from both sides.0.20x210.Simplify.

Finally, we isolate x from 0.20x by dividing both sides of the inequality by 0.20:

0.20x0.202100.20Divide both sides by 0.20.x1050.Simplify.

With at most $335 to spend, you can travel at most 1050 miles.

We started with the inequality 0.20x+125335 and obtained the inequality x1050 in the final step. These inequalities have the same solution set, namely, {x|x1050}. Inequalities such as these, with the same solution set, are said to be equivalent.

We isolated x from 0.20x by dividing both sides of 0.20x210 by 0.20, a positive number. Let’s see what happens if we divide both sides of an inequality by a negative number. Consider the inequality 10<14. Divide 10 and 14 by 2:

102=5and142=7.

Because 5 lies to the right of 7 on the number line, 5 is greater than 7:

5>7.

Notice that the direction of the inequality symbol is reversed:

A two way arrow between 10 is less than 14 and negative 5 is greater than negative 7. A note reads, dividing by negative 2 changes the direction of the inequality symbol.

In general, when we multiply or divide both sides of an inequality by a negative number, the direction of the inequality symbol is reversed. When we reverse the direction of the inequality symbol, we say that we change the sense of the inequality.

We can isolate a variable in a linear inequality in the same way we isolate a variable in a linear equation. The properties on the next page are used to create equivalent inequalities.

Properties of Inequalities

Property The Property in Words Example

The Addition Property of Inequality

If a<b, then a+c<b+c.

If a<b, then ac<bc.

If the same quantity is added to or subtracted from both sides of an inequality, the resulting inequality is equivalent to the original one.

2x+3<7

Subtract 3:

2x+33<73.

Simplify:

2x<4.

The Positive Multiplication Property of Inequality

If a<b and c is positive, then ac<bc.

If a<b and c is positive, then ac<bc.

If we multiply or divide both sides of an inequality by the same positive quantity, the resulting inequality is equivalent to the original one.

2x<4

Divide by 2:

2x2<42.

Simplify:

x<2.

The Negative Multiplication Property of Inequality

If a<b and c is negative, then ac>bc.

If a<b and c is negative, then ac>bc.

If we multiply or divide both sides of an inequality by the same negative quantity and reverse the direction of the inequality symbol, the resulting inequality is equivalent to the original one.

4x<20

Divide by 4 and change the sense of the inequality:

4x4>204.

Simplify:

x>5.

Example 3 Solving a Linear Inequality

Solve and graph the solution set on a number line:

32x11.

Solution

32x11This is the given inequality.32x3113Subtract 3 from both sides.2x8Simplify.2x282Divide both sides by 2  andchange the sense of the inequality.x4Simplify.

The solution set of 32x11, or equivalently x4, consists of all real numbers that are greater than or equal to 4, expressed as {x|x4} in set-builder notation. The interval notation for this solution set is [4, ). The graph of the solution set is shown as follows:

A number line ranging from negative 5 to 5 is shaded right to a left bracket at 4.

Check Point 3

  • Solve and graph the solution set on a number line:

    23x5.

Example 4 Solving a Linear Inequality

Solve and graph the solution set on a number line:

2x4>x+5.

Solution

  1. Step 1 SIMPLIFY EACH SIDE. Because each side is already simplified, we can skip this step.

  2. Step 2 COLLECT VARIABLE TERMS ON ONE SIDE AND CONSTANT TERMS ON THE OTHER SIDE. We will collect variable terms of 2x4>x+5 on the left and constant terms on the right.

    2x4>x+5This  is  the  given  inequlity.2x4x>x+5xSubtract  x  from  both  sides.3x4>5Simplify.3x4+4>5+4Add  4  to  both  sides.3x>9Simplify.
  3. Step 3 ISOLATE THE VARIABLE AND SOLVE. We isolate the variable, x, by dividing both sides by 3. Because we are dividing by a negative number, we must reverse the inequality symbol.

    3x3<93Divide both sides by 3 and   change the sense of the inequality.x<3   Simplify.
  4. Step 4 EXPRESS THE SOLUTION SET IN SET-BUILDER OR INTERVAL NOTATION AND GRAPH THE SET ON A NUMBER LINE. The solution set consists of all real numbers that are less than 3, expressed in set-builder notation as {x|x<3}. The interval notation for this solution set is (, 3). The graph of the solution set is shown as follows:

    A number line ranging from negative 5 to 5 is shaded left to a right parenthesis at negative 3.

Check Point 4

  • Solve and graph the solution set on a number line: 3x+1>7x15.

If an inequality contains fractions with constants in the denominators, begin by multiplying both sides by the least common denominator. This will clear the inequality of fractions.

Example 5 Solving a Linear Inequality Containing Fractions

Solve and graph the solution set on a number line:

x+34x23+14.

Solution

The denominators are 4, 3, and 4. The least common denominator is 12. We begin by multiplying both sides of the inequality by 12.

The image shows the steps to solve an inequality.

Now that the fractions have been cleared, we follow the four steps that we used in the previous example.

  1. Step 1 SIMPLIFY EACH SIDE.

    3(x+3)4(x2)+3This is the inequality with the fractions cleared.3x+94x8+3Use the distributive property.3x+94x5Simplify.
  2. Step 2 COLLECT VARIABLE TERMS ON ONE SIDE AND CONSTANT TERMS ON THE OTHER SIDE. We will collect variable terms of 3x+94x5 on the left and constant terms on the right.

    3x+94x4x54xSubtract 4x from both sides.x+95Simplify.x+9959Subtract 9 from both sides.x14Simplify.
  3. Step 3 ISOLATE THE VARIABLE AND SOLVE. To isolate x, we must eliminate the negative sign in front of the x. Because x means 1x, we can do this by multiplying (or dividing) both sides of the inequality by 1. We are multiplying by a negative number. Thus, we must reverse the direction of the inequality symbol.

    (1)(x)(1)(14) Multiply both sides by 1 and change the sense of the inequality.x14Simplify.
  4. Step 4 EXPRESS THE SOLUTION SET IN SET-BUILDER OR INTERVAL NOTATION AND GRAPH THE SET ON A NUMBER LINE. The solution set consists of all real numbers that are less than or equal to 14, expressed in set-builder notation as {x|x14}. The interval notation for this solution set is (, 14]. The graph of the solution set is shown as follows:

    A number line ranging from 5 to 15 is shaded left to a right bracket at 14.

Check Point 5

  • Solve and graph the solution set on a number line:

    x42x23+56.
Objective 4: Solve compound inequalities

Solving Compound Inequalities

  1. Objective 4Solve compound inequalities.

Watch Video

We now consider two inequalities such as

3<2x+1and2x+13,

expressed as a compound inequality

3<2x+13.

The word and does not appear when the inequality is written in the shorter form, although intersection is implied. The shorter form enables us to solve both inequalities at once. By performing each operation on all three parts of the inequality, our goal is to isolate x in the middle.

Example 6 Solving a Compound Inequality

Solve and graph the solution set on a number line:

3<2x+13.

Solution

We would like to isolate x in the middle. We can do this by first subtracting 1 from all three parts of the compound inequality. Then we isolate x from 2x by dividing all three parts of the inequality by 2.

3<2x+13This is the given inequality.31<2x+1131Subtract 1 from all three parts.4<2x2Simplify.42<2x222Divide each part by 2.2<x1Simplify.

The solution set consists of all real numbers greater than 2 and less than or equal to 1, represented by {x|2<x1} in set-builder notation and (2, 1] in interval notation. The graph is shown as follows:

A number line ranging from negative 5 to 5 is shaded from a left parenthesis at negative 2 to a right bracket at 1.

Check Point 6

  • Solve and graph the solution set on a number line: 12x+3<11.

Objective 5: Solve absolute value inequalities

Solving Inequalities with Absolute Value

  1. Objective 5Solve absolute value inequalities.

Watch Video

We know that |x| describes the distance of x from zero on a real number line. We can use this geometric interpretation to solve an inequality such as

|x|<2.

This means that the distance of x from 0 is less than 2, as shown in Figure P.22. The interval shows values of x that lie less than 2 units from 0. Thus, x can lie between 2 and 2. That is, x is greater than 2 and less than 2. We write (2, 2) or {x|2<x<2}.

Figure P.22 |x|<2, so 2<x<2.

A number line ranging from negative 4 to 4 is shaded from a left parenthesis at negative 2 and a right parenthesis at 2.

Some absolute value inequalities use the “greater than” symbol. For example, |x|>2 means that the distance of x from 0 is greater than 2, as shown in Figure P.23. Thus, x can be less than 2 or greater than 2. We write x<2 or x>2. This can be expressed in interval notation as (, 2)  (2, ).

Figure P.23 |x|>2, so x<2 or x>2.

A number line ranging from negative 4 to 4 is shaded left to a right parenthesis at negative 2 and right to a left parenthesis at 2.

These observations suggest the following principles for solving inequalities with absolute value.

Solving an Absolute Value Inequality

If u is an algebraic expression and c is a positive number,

  1. The solutions of |u|< c are the numbers that satisfy c<u<c.

  2. The solutions of |u|> c are the numbers that satisfy u<c or u>c.

These rules are valid if < is replaced by ≤ and > is replaced by ≥.

Example 7 Solving an Absolute Value Inequality

Solve and graph the solution set on a number line: |x4|<3.

Solution

We rewrite the inequality without absolute value bars.

Vertical bar u vertical bar is less than c means negative c is less than u is less than c. Vertical bar x minus 4 vertical bar is less than 3 means negative 3 is less than x minus 4 is less than 3.

We solve the compound inequality by adding 4 to all three parts.

3<x4<33+4<x4+4<3+41<x<7

The solution set of |x4|<3 consists of all real numbers greater than 1 and less than 7, denoted by {x|1 < x < 7} or (1, 7). The graph of the solution set is shown as follows:

A number line ranging from negative 2 to 8 is shaded from a left parenthesis at 1 to a right parenthesis at 7.

Check Point 7

  • Solve and graph the solution set on a number line: |x2|<5.

Example 8 Solving an Absolute Value Inequality

Solve and graph the solution set on a number line: 2|3x+5|+713.

Solution

An image shows steps to solve an inequality.

The solution set is {x|5x53} in set-builder notation and [5, 53] in interval notation. The graph is shown as follows:

A number line ranging from negative 7 to 3 is shaded from a left bracket at negative 5 to a right bracket at 1 and 2 thirds.

Check Point 8

  • Solve and graph the solution set on a number line: 3|5x2|+2019.

Example 9 Solving an Absolute Value Inequality

Solve and graph the solution set on a number line: 7 < |52x|.

Solution

We begin by expressing the inequality with the absolute value expression on the left side:

Vertical bar 5 minus 2 x vertical bar is greater than 7. A note reads, c is less than the vertical bar u vertical bar, means the same thing as the vertical bar u vertical bar is greater than c. In both cases comma the inequality symbol points to c.

We rewrite this inequality without absolute value bars.

Vertcal bar u vertical bar is greater than c means u is less than negative c or u is greater than c. Vertical bar of 5 minus 2 x vertical bar is greater than 7 means 5 minus 2 x is less than negative 7 or 5 minus 2 x is greater than 7.

Because |52x|>7 means 52x <7 or 52x>7, we solve 52x <7 and 52x > 7 separately. Then we take the union of their solution sets.

52x<7or52x>7These are the inequalities withoutabsolute value bars.552x<75552x>75 Subtract 5 from both sides.2x<122x>2 Simplify.2x2>1222x2<22Divide both sides by 2 and changethe sense of each inequality.x>6x<1 Simplify.

The solution set consists of all numbers that are less than 1 or greater than 6. The solution set is {x|x<1 orx>6}, or, in interval notation (, 1)(6, ). The graph of the solution set is shown as follows:

A number line ranging from negative 3 to 8 is shaded left to a right parenthesis at negative 1 and right to a left parenthesis at 6.

Check Point 9

  • Solve and graph the solution set on a number line: 18<|63x|.

Applications

In Section P.8, we solved equations to determine when two different pricing options would result in the same cost. With inequalities, we can look at the same situations and ask when one of the pricing options results in a lower cost than the other.

Our next example shows how to use an inequality to select the better deal between our two options for paying the bridge toll from Example 3 in Section P.8. We use our strategy for solving word problems, modeling the verbal conditions of the problem with a linear inequality.

Example 10 Selecting the Better Deal

You still have two options for paying the toll so that you can get to the beach. The first option requires purchasing a transponder for $20; with the transponder, you pay a reduced toll of $3.25 each time you cross the bridge. Your second option is toll-by-plate; with this option, you pay the full toll of $4.25 and a $3 administrative fee each time you cross the bridge, for a total of $7.25 for each crossing. Find the number of times you would need to cross the bridge for the transponder option to be the better deal.

Solution

  1. Step 1 LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We are looking for the number of times you must cross the bridge to make the transponder option the better deal. Thus,

    let x=the number of times you cross the bridge.
  2. Step 2 REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. We are not asked to find another quantity, so we can skip this step.

  3. Step 3 WRITE AN INEQUALITY IN x THAT MODELS THE CONDITIONS. The transponder is a better deal than toll-by-plate if the total cost with the transponder is less than the total cost of toll-by-plate.

    The Image shows the inequality in x that models the conditions.
  4. Step 4 SOLVE THE INEQUALITY AND ANSWER THE QUESTION.

    20+3.25x<7.25xThis is the inequality that models theverbal conditions.20+3.25x7.25x<7.25x7.25x Subtract 7.25x from both sides.204x<0 Simplify.204x20<020 Subtract 20 from both sides.4x<20 Simplify.4x4>204Divide both sides by 4 and changethe sense of the inequality.x>5 Simplify.

    Thus, crossing the bridge more than five times makes the transponder option the better deal.

  5. Step 5 CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. One way to do this is to take a number of crossings greater than five and see if the transponder option is the better deal. Suppose that you cross the bridge six times.

    Cost, in dollars, with transponder=20+3.25(6)=39.50Cost, in dollars, of toll-by-plate=7.25(6)=43.50

    The cost with the transponder is lower, making this option the better deal.

Check Point 10

  • You drive up to a toll plaza and find booths with attendants, and you can pay the toll by cash or credit card. With this option, the toll is $5 each time you cross the bridge. The attendant gives you the option of buying a bar-coded decal for $25; with the decal, you get 25% off the normal toll of $5 for each crossing. Find the number of times you would need to cross the bridge to make the decal option the better deal.

Objective 5: Solve absolute value inequalities

Solving Inequalities with Absolute Value

  1. Objective 5Solve absolute value inequalities.

Watch Video

We know that |x| describes the distance of x from zero on a real number line. We can use this geometric interpretation to solve an inequality such as

|x|<2.

This means that the distance of x from 0 is less than 2, as shown in Figure P.22. The interval shows values of x that lie less than 2 units from 0. Thus, x can lie between 2 and 2. That is, x is greater than 2 and less than 2. We write (2, 2) or {x|2<x<2}.

Figure P.22 |x|<2, so 2<x<2.

A number line ranging from negative 4 to 4 is shaded from a left parenthesis at negative 2 and a right parenthesis at 2.

Some absolute value inequalities use the “greater than” symbol. For example, |x|>2 means that the distance of x from 0 is greater than 2, as shown in Figure P.23. Thus, x can be less than 2 or greater than 2. We write x<2 or x>2. This can be expressed in interval notation as (, 2)  (2, ).

Figure P.23 |x|>2, so x<2 or x>2.

A number line ranging from negative 4 to 4 is shaded left to a right parenthesis at negative 2 and right to a left parenthesis at 2.

These observations suggest the following principles for solving inequalities with absolute value.

Solving an Absolute Value Inequality

If u is an algebraic expression and c is a positive number,

  1. The solutions of |u|< c are the numbers that satisfy c<u<c.

  2. The solutions of |u|> c are the numbers that satisfy u<c or u>c.

These rules are valid if < is replaced by ≤ and > is replaced by ≥.

Example 7 Solving an Absolute Value Inequality

Solve and graph the solution set on a number line: |x4|<3.

Solution

We rewrite the inequality without absolute value bars.

Vertical bar u vertical bar is less than c means negative c is less than u is less than c. Vertical bar x minus 4 vertical bar is less than 3 means negative 3 is less than x minus 4 is less than 3.

We solve the compound inequality by adding 4 to all three parts.

3<x4<33+4<x4+4<3+41<x<7

The solution set of |x4|<3 consists of all real numbers greater than 1 and less than 7, denoted by {x|1 < x < 7} or (1, 7). The graph of the solution set is shown as follows:

A number line ranging from negative 2 to 8 is shaded from a left parenthesis at 1 to a right parenthesis at 7.

Check Point 7

  • Solve and graph the solution set on a number line: |x2|<5.

Example 8 Solving an Absolute Value Inequality

Solve and graph the solution set on a number line: 2|3x+5|+713.

Solution

An image shows steps to solve an inequality.

The solution set is {x|5x53} in set-builder notation and [5, 53] in interval notation. The graph is shown as follows:

A number line ranging from negative 7 to 3 is shaded from a left bracket at negative 5 to a right bracket at 1 and 2 thirds.

Check Point 8

  • Solve and graph the solution set on a number line: 3|5x2|+2019.

Example 9 Solving an Absolute Value Inequality

Solve and graph the solution set on a number line: 7 < |52x|.

Solution

We begin by expressing the inequality with the absolute value expression on the left side:

Vertical bar 5 minus 2 x vertical bar is greater than 7. A note reads, c is less than the vertical bar u vertical bar, means the same thing as the vertical bar u vertical bar is greater than c. In both cases comma the inequality symbol points to c.

We rewrite this inequality without absolute value bars.

Vertcal bar u vertical bar is greater than c means u is less than negative c or u is greater than c. Vertical bar of 5 minus 2 x vertical bar is greater than 7 means 5 minus 2 x is less than negative 7 or 5 minus 2 x is greater than 7.

Because |52x|>7 means 52x <7 or 52x>7, we solve 52x <7 and 52x > 7 separately. Then we take the union of their solution sets.

52x<7or52x>7These are the inequalities withoutabsolute value bars.552x<75552x>75 Subtract 5 from both sides.2x<122x>2 Simplify.2x2>1222x2<22Divide both sides by 2 and changethe sense of each inequality.x>6x<1 Simplify.

The solution set consists of all numbers that are less than 1 or greater than 6. The solution set is {x|x<1 orx>6}, or, in interval notation (, 1)(6, ). The graph of the solution set is shown as follows:

A number line ranging from negative 3 to 8 is shaded left to a right parenthesis at negative 1 and right to a left parenthesis at 6.

Check Point 9

  • Solve and graph the solution set on a number line: 18<|63x|.

Applications

In Section P.8, we solved equations to determine when two different pricing options would result in the same cost. With inequalities, we can look at the same situations and ask when one of the pricing options results in a lower cost than the other.

Our next example shows how to use an inequality to select the better deal between our two options for paying the bridge toll from Example 3 in Section P.8. We use our strategy for solving word problems, modeling the verbal conditions of the problem with a linear inequality.

Example 10 Selecting the Better Deal

You still have two options for paying the toll so that you can get to the beach. The first option requires purchasing a transponder for $20; with the transponder, you pay a reduced toll of $3.25 each time you cross the bridge. Your second option is toll-by-plate; with this option, you pay the full toll of $4.25 and a $3 administrative fee each time you cross the bridge, for a total of $7.25 for each crossing. Find the number of times you would need to cross the bridge for the transponder option to be the better deal.

Solution

  1. Step 1 LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We are looking for the number of times you must cross the bridge to make the transponder option the better deal. Thus,

    let x=the number of times you cross the bridge.
  2. Step 2 REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. We are not asked to find another quantity, so we can skip this step.

  3. Step 3 WRITE AN INEQUALITY IN x THAT MODELS THE CONDITIONS. The transponder is a better deal than toll-by-plate if the total cost with the transponder is less than the total cost of toll-by-plate.

    The Image shows the inequality in x that models the conditions.
  4. Step 4 SOLVE THE INEQUALITY AND ANSWER THE QUESTION.

    20+3.25x<7.25xThis is the inequality that models theverbal conditions.20+3.25x7.25x<7.25x7.25x Subtract 7.25x from both sides.204x<0 Simplify.204x20<020 Subtract 20 from both sides.4x<20 Simplify.4x4>204Divide both sides by 4 and changethe sense of the inequality.x>5 Simplify.

    Thus, crossing the bridge more than five times makes the transponder option the better deal.

  5. Step 5 CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. One way to do this is to take a number of crossings greater than five and see if the transponder option is the better deal. Suppose that you cross the bridge six times.

    Cost, in dollars, with transponder=20+3.25(6)=39.50Cost, in dollars, of toll-by-plate=7.25(6)=43.50

    The cost with the transponder is lower, making this option the better deal.

Check Point 10

  • You drive up to a toll plaza and find booths with attendants, and you can pay the toll by cash or credit card. With this option, the toll is $5 each time you cross the bridge. The attendant gives you the option of buying a bar-coded decal for $25; with the decal, you get 25% off the normal toll of $5 for each crossing. Find the number of times you would need to cross the bridge to make the decal option the better deal.

Objective 5: Solve absolute value inequalities

Solving Inequalities with Absolute Value

  1. Objective 5Solve absolute value inequalities.

Watch Video

We know that |x| describes the distance of x from zero on a real number line. We can use this geometric interpretation to solve an inequality such as

|x|<2.

This means that the distance of x from 0 is less than 2, as shown in Figure P.22. The interval shows values of x that lie less than 2 units from 0. Thus, x can lie between 2 and 2. That is, x is greater than 2 and less than 2. We write (2, 2) or {x|2<x<2}.

Figure P.22 |x|<2, so 2<x<2.

A number line ranging from negative 4 to 4 is shaded from a left parenthesis at negative 2 and a right parenthesis at 2.

Some absolute value inequalities use the “greater than” symbol. For example, |x|>2 means that the distance of x from 0 is greater than 2, as shown in Figure P.23. Thus, x can be less than 2 or greater than 2. We write x<2 or x>2. This can be expressed in interval notation as (, 2)  (2, ).

Figure P.23 |x|>2, so x<2 or x>2.

A number line ranging from negative 4 to 4 is shaded left to a right parenthesis at negative 2 and right to a left parenthesis at 2.

These observations suggest the following principles for solving inequalities with absolute value.

Solving an Absolute Value Inequality

If u is an algebraic expression and c is a positive number,

  1. The solutions of |u|< c are the numbers that satisfy c<u<c.

  2. The solutions of |u|> c are the numbers that satisfy u<c or u>c.

These rules are valid if < is replaced by ≤ and > is replaced by ≥.

Example 7 Solving an Absolute Value Inequality

Solve and graph the solution set on a number line: |x4|<3.

Solution

We rewrite the inequality without absolute value bars.

Vertical bar u vertical bar is less than c means negative c is less than u is less than c. Vertical bar x minus 4 vertical bar is less than 3 means negative 3 is less than x minus 4 is less than 3.

We solve the compound inequality by adding 4 to all three parts.

3<x4<33+4<x4+4<3+41<x<7

The solution set of |x4|<3 consists of all real numbers greater than 1 and less than 7, denoted by {x|1 < x < 7} or (1, 7). The graph of the solution set is shown as follows:

A number line ranging from negative 2 to 8 is shaded from a left parenthesis at 1 to a right parenthesis at 7.

Check Point 7

  • Solve and graph the solution set on a number line: |x2|<5.

Example 8 Solving an Absolute Value Inequality

Solve and graph the solution set on a number line: 2|3x+5|+713.

Solution

An image shows steps to solve an inequality.

The solution set is {x|5x53} in set-builder notation and [5, 53] in interval notation. The graph is shown as follows:

A number line ranging from negative 7 to 3 is shaded from a left bracket at negative 5 to a right bracket at 1 and 2 thirds.

Check Point 8

  • Solve and graph the solution set on a number line: 3|5x2|+2019.

Example 9 Solving an Absolute Value Inequality

Solve and graph the solution set on a number line: 7 < |52x|.

Solution

We begin by expressing the inequality with the absolute value expression on the left side:

Vertical bar 5 minus 2 x vertical bar is greater than 7. A note reads, c is less than the vertical bar u vertical bar, means the same thing as the vertical bar u vertical bar is greater than c. In both cases comma the inequality symbol points to c.

We rewrite this inequality without absolute value bars.

Vertcal bar u vertical bar is greater than c means u is less than negative c or u is greater than c. Vertical bar of 5 minus 2 x vertical bar is greater than 7 means 5 minus 2 x is less than negative 7 or 5 minus 2 x is greater than 7.

Because |52x|>7 means 52x <7 or 52x>7, we solve 52x <7 and 52x > 7 separately. Then we take the union of their solution sets.

52x<7or52x>7These are the inequalities withoutabsolute value bars.552x<75552x>75 Subtract 5 from both sides.2x<122x>2 Simplify.2x2>1222x2<22Divide both sides by 2 and changethe sense of each inequality.x>6x<1 Simplify.

The solution set consists of all numbers that are less than 1 or greater than 6. The solution set is {x|x<1 orx>6}, or, in interval notation (, 1)(6, ). The graph of the solution set is shown as follows:

A number line ranging from negative 3 to 8 is shaded left to a right parenthesis at negative 1 and right to a left parenthesis at 6.

Check Point 9

  • Solve and graph the solution set on a number line: 18<|63x|.

Applications

In Section P.8, we solved equations to determine when two different pricing options would result in the same cost. With inequalities, we can look at the same situations and ask when one of the pricing options results in a lower cost than the other.

Our next example shows how to use an inequality to select the better deal between our two options for paying the bridge toll from Example 3 in Section P.8. We use our strategy for solving word problems, modeling the verbal conditions of the problem with a linear inequality.

Example 10 Selecting the Better Deal

You still have two options for paying the toll so that you can get to the beach. The first option requires purchasing a transponder for $20; with the transponder, you pay a reduced toll of $3.25 each time you cross the bridge. Your second option is toll-by-plate; with this option, you pay the full toll of $4.25 and a $3 administrative fee each time you cross the bridge, for a total of $7.25 for each crossing. Find the number of times you would need to cross the bridge for the transponder option to be the better deal.

Solution

  1. Step 1 LET x REPRESENT ONE OF THE UNKNOWN QUANTITIES. We are looking for the number of times you must cross the bridge to make the transponder option the better deal. Thus,

    let x=the number of times you cross the bridge.
  2. Step 2 REPRESENT OTHER UNKNOWN QUANTITIES IN TERMS OF x. We are not asked to find another quantity, so we can skip this step.

  3. Step 3 WRITE AN INEQUALITY IN x THAT MODELS THE CONDITIONS. The transponder is a better deal than toll-by-plate if the total cost with the transponder is less than the total cost of toll-by-plate.

    The Image shows the inequality in x that models the conditions.
  4. Step 4 SOLVE THE INEQUALITY AND ANSWER THE QUESTION.

    20+3.25x<7.25xThis is the inequality that models theverbal conditions.20+3.25x7.25x<7.25x7.25x Subtract 7.25x from both sides.204x<0 Simplify.204x20<020 Subtract 20 from both sides.4x<20 Simplify.4x4>204Divide both sides by 4 and changethe sense of the inequality.x>5 Simplify.

    Thus, crossing the bridge more than five times makes the transponder option the better deal.

  5. Step 5 CHECK THE PROPOSED SOLUTION IN THE ORIGINAL WORDING OF THE PROBLEM. One way to do this is to take a number of crossings greater than five and see if the transponder option is the better deal. Suppose that you cross the bridge six times.

    Cost, in dollars, with transponder=20+3.25(6)=39.50Cost, in dollars, of toll-by-plate=7.25(6)=43.50

    The cost with the transponder is lower, making this option the better deal.

Check Point 10

  • You drive up to a toll plaza and find booths with attendants, and you can pay the toll by cash or credit card. With this option, the toll is $5 each time you cross the bridge. The attendant gives you the option of buying a bar-coded decal for $25; with the decal, you get 25% off the normal toll of $5 for each crossing. Find the number of times you would need to cross the bridge to make the decal option the better deal.

P.9: Exercise Set

P.9 Exercise Set

Practice Exercises

In Exercises 114, express each interval in set-builder notation and graph the interval on a number line.

  1. 1. (1, 6]

  2. 2. (2, 4]

  3. 3. [5, 2)

  4. 4. [4, 3)

  5. 5. [3, 1]

  6. 6. [2, 5]

  7. 7. (2, )

  8. 8. (3, )

  9. 9. [3, )

  10. 10. [5, )

  11. 11. (, 3)

  12. 12. (, 2)

  13. 13. (, 5.5)

  14. 14. (, 3.5]

In Exercises 1526, use graphs to find each set.

  1. 15. (3, 0)[1, 2]

  2. 16. (4, 0)[2, 1]

  3. 17. (3, 0)[1, 2]

  4. 18. (4, 0)[2, 1]

  5. 19. (, 5)[1, 8)

  6. 20. (, 6)[2, 9)

  7. 21. (, 5)[1, 8)

  8. 22. (, 6)[2, 9)

  9. 23. [3, )(6, )

  10. 24. [2, )(4, )

  11. 25. [3, )(6, )

  12. 26. [2, )(4, )

In all exercises, use interval notation to express solution sets and graph each solution set on a number line.

In Exercises 2748, solve each linear inequality.

  1. 27. 5x+11<26

  2. 28. 2x+5<17

  3. 29. 3x713

  4. 30. 8x214

  5. 31. 9x36

  6. 32. 5x30

  7. 33. 8x113x13

  8. 34. 18x+4512x8

  9. 35. 4(x+1)+23x+6

  10. 36. 8x+3>3(2x+1)+x+5

  11. 37. 2x11<3(x+2)

  12. 38. 4(x+2)>3x+20

  13. 39. 1(x+3)42x

  14. 40. 5(3x)3x1

  15. 41. x432x2+1

  16. 42. 3x10+115x10

  17. 43. 1x2>4

  18. 44. 745 x<35

  19. 45. x46x29+518

  20. 46. 4x36+22x112

  21. 47. 3[3(x+5)+8x+7]+5[3(x6)2(3x5)]<2(4x+3)

  22. 48. 5[3(23x)2(5x)]6[5(x2)2(4x3)]<3x+19

In Exercises 4956, solve each compound inequality.

  1. 49. 6<x+3<8

  2. 50. 7<x+5<11

  3. 51. 3x2<1

  4. 52. 6<x41

  5. 53. 11<2x15

  6. 54. 34x3<19

  7. 55. 323 x5<1

  8. 56. 612 x4<3

In Exercises 5792, solve each absolute value inequality.

  1. 57. |x|<3

  2. 58. |x|<5

  3. 59. |x1|2

  4. 60. |x+3|4

  5. 61. |2x6|<8

  6. 62. |3x+5|<17

  7. 63. |2(x1)+4|8

  8. 64. |3(x1)+2|20

  9. 65. |2x+63|<2

  10. 66. |3(x1)4|<6

  11. 67. |x|>3

  12. 68. |x|>5

  13. 69. |x1|2

  14. 70. |x+3|4

  15. 71. |3x8|>7

  16. 72. |5x2|>13

  17. 73. |2x+24|2

  18. 74. |3x39|1

  19. 75. |323 x|>5

  20. 76. |334 x|>9

  21. 77. 3|x1|+28

  22. 78. 5|2x+1|39

  23. 79. 2|x4|4

  24. 80. 3|x+7|27

  25. 81. 4|1x|<16

  26. 82. 2|5x|<6

  27. 83. 3|2x1|

  28. 84. 9|4x+7|

  29. 85. 5>|4x|

  30. 86. 2>|11x|

  31. 87. 1<|23x|

  32. 88. 4<|2x|

  33. 89. 12<|2x+67|+37

  34. 90. 1<|x113|+73

  35. 91. 4+|3x3|9

  36. 92. |2x2|11

Practice PLUS

In Exercises 93100, use interval notation to represent all values of x satisfying the given conditions.

  1. 93. y1=x2+3, y2=x3+52, and y1y2.

  2. 94. y1=23 (6x9)+4, y2=5x+1, and y1>y2.

  3. 95. y=1(x+3)+2x and y is at least 4.

  4. 96. y=2x11+3(x+2) and y is at most 0.

  5. 97. y=|3x4|+2 and y<8.

  6. 98. y=|2x5|+1 and y>9.

  7. 99. y=7|x2+2| and y is at most 4.

  8. 100. y=8|5x+3| and y is at least 6.

  9. 101. When 3 times a number is subtracted from 4, the absolute value of the difference is at least 5. Use interval notation to express the set of all numbers that satisfy this condition.

  10. 102. When 4 times a number is subtracted from 5, the absolute value of the difference is at most 13. Use interval notation to express the set of all numbers that satisfy this condition.

Application Exercises

The graphs show that the three components of love, namely, passion, intimacy, and commitment, progress differently over time. Passion peaks early in a relationship and then declines. By contrast, intimacy and commitment build gradually. Use the graphs to solve Exercises 103110.

A line graph depicts the course of love over time.

Source: R. J. Sternberg. A Triangular Theory of Love, Psychological Review, 93, 119–135

  1. 103. Use interval notation to write an inequality that expresses for which years in a relationship intimacy is greater than commitment.

  2. 104. Use interval notation to write an inequality that expresses for which years in a relationship passion is greater than or equal to intimacy.

  3. 105. What is the relationship between passion and intimacy on the interval [5, 7)?

  4. 106. What is the relationship between intimacy and commitment on the interval [4, 7)?

  5. 107. What is the relationship between passion and commitment for {x|6<x<8}?

  6. 108. What is the relationship between passion and commitment for {x|7<x<9}?

  7. 109. What is the maximum level of intensity for passion? After how many years in a relationship does this occur?

  8. 110. After approximately how many years do levels of intensity for commitment exceed the maximum level of intensity for passion?

In more U.S. marriages, spouses have different faiths. The bar graph shows the percentage of households with an interfaith marriage in 1988 and 2012. Also shown is the percentage of households in which a person of faith is married to someone with no religion.

A bar chart shows percentage of U.S. households in which married couples do not share the same faith.

Source: General Social Survey, University of Chicago

The formula

I=14x+26

models the percentage of U.S. households with an interfaith marriage, I, x years after 1988. The formula

N=14x+6

models the percentage of U.S. households in which a person of faith is married to someone with no religion, N, x years after 1988.

Use these models to solve Exercises 111112.

  1. 111.

    1. In which years will more than 33% of U.S. households have an interfaith marriage?

    2. In which years will more than 14% of U.S. households have a person of faith married to someone with no religion?

    3. Based on your answers to parts (a) and (b), in which years will more than 33% of households have an interfaith marriage and more than 14% have a faith/no religion marriage?

    4. Based on your answers to parts (a) and (b), in which years will more than 33% of households have an interfaith marriage or more than 14% have a faith/no religion marriage?

  2. 112.

    1. In which years will more than 34% of U.S. households have an interfaith marriage?

    2. In which years will more than 15% of U.S. households have a person of faith married to someone with no religion?

    3. Based on your answers to parts (a) and (b), in which years will more than 34% of households have an interfaith marriage and more than 15% have a faith/no religion marriage?

    4. Based on your answers to parts (a) and (b), in which years will more than 34% of households have an interfaith marriage or more than 15% have a faith/no religion marriage?

  3. 113. The formula for converting Fahrenheit temperature, F, to Celsius temperature, C, is

    C=59 (F32).

    If Celsius temperature ranges from 15° to 35°, inclusive, what is the range for the Fahrenheit temperature? Use interval notation to express this range.

  4. 114. The formula for converting Celsius temperature, C, to Fahrenheit temperature, F, is

    F = 95 C+ 32.

    If Fahrenheit temperature ranges from 41° to 50°, inclusive, what is the range for Celsius temperature? Use interval notation to express this range.

  5. 115. If a coin is tossed 100 times, we would expect approximately 50 of the outcomes to be heads. It can be demonstrated that a coin is unfair if h, the number of outcomes that result in heads, satisfies |h505|1.645. Describe the number of outcomes that determine an unfair coin that is tossed 100 times.

In Exercises 116127, use the strategy for solving word problems, modeling the verbal conditions of the problem with a linear inequality.

  1. 116. You see a sign on a small moving truck that reads “Rent me for $20 a day.*” Oh, there’s an asterisk: There is an additional charge of $0.80 per mile. When you go online to reserve the truck, you are offered an unlimited mileage option for $60 a day. How many miles would you have to drive the truck in a day to make the unlimited mileage option the better deal?

  2. 117. A transponder for a toll bridge costs $27.50. With the transponder, the toll is $5 each time you cross the bridge. The only other option is toll-by-plate, for which the toll is $6.25 each time you cross the bridge with an additional administrative fee of $1.25 for each crossing. For how many bridge crossings is toll-by-plate the better option?

  3. 118. An electronic pass for a toll road costs $30. The toll is normally $5.00 but is reduced by 30% for people who have purchased the electronic pass. Determine the number of times the road must be used so that the electronic pass option is the better deal.

  4. 119. A city commission has proposed two tax bills. The first bill requires that a homeowner pay $1200 plus 0.5% of the assessed home value in taxes. The second bill requires taxes of $300 plus 0.9% of the assessed home value. What price range of home assessment would make the first bill a better deal?

  5. 120. A company designs and sells greeting cards. The weekly fixed cost is $10,000 and it costs $0.40 to create each card. The selling price is $2.00 per card. How many greeting cards must be designed and sold each week for the company to generate a profit?

  6. 121. A company manufactures and sells personalized stationery. The weekly fixed cost is $3000 and it costs $3.00 to produce each package of stationery. The selling price is $5.50 per package. How many packages of stationery must be produced and sold each week for the company to generate a profit?

  7. 122. An elevator at a construction site has a maximum capacity of 2800 pounds. If the elevator operator weighs 265 pounds and each cement bag weighs 65 pounds, how many bags of cement can be safely lifted on the elevator in one trip?

  8. 123. An elevator at a construction site has a maximum capacity of 3000 pounds. If the elevator operator weighs 245 pounds and each cement bag weighs 95 pounds, how many bags of cement can be safely lifted on the elevator in one trip?

  9. 124. To earn an A in a course, you must have a final average of at least 90%. On the first four examinations, you have grades of 86%, 88%, 92%, and 84%. If the final examination counts as two grades, what must you get on the final to earn an A in the course?

  10. 125. On two examinations, you have grades of 86 and 88. There is an optional final examination, which counts as one grade. You decide to take the final in order to get a course grade of A, meaning a final average of at least 90.

    1. What must you get on the final to earn an A in the course?

    2. By taking the final, if you do poorly, you might risk the B that you have in the course based on the first two exam grades. If your final average is less than 80, you will lose your B in the course. Describe the grades on the final that will cause this to happen.

  11. 126. Parts for an automobile repair cost $254. The mechanic charges $65 per hour. If you receive an estimate for at least $351.50 and at most $481.50 for fixing the car, what is the time interval that the mechanic will be working on the job?

  12. 127. The toll to a bridge is $3.00. A three-month pass costs $7.50 and reduces the toll to $0.50. A six-month pass costs $30 and permits crossing the bridge for no additional fee. How many crossings per three-month period does it take for the three-month pass to be the best deal?

Explaining the Concepts

  1. 128. When graphing the solutions of an inequality, what does a parenthesis signify? What does a square bracket signify?

  2. 129. Describe ways in which solving a linear inequality is similar to solving a linear equation.

  3. 130. Describe ways in which solving a linear inequality is different from solving a linear equation.

  4. 131. What is a compound inequality and how is it solved?

  5. 132. Describe how to solve an absolute value inequality involving the symbol <. Give an example.

  6. 133. Describe how to solve an absolute value inequality involving the symbol >. Give an example.

  7. 134. Explain why |x|<4 has no solution.

  8. 135. Describe the solution set of |x|>4.

Critical Thinking Exercises

Make Sense? In Exercises 136139, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 136. I prefer interval notation over set-builder notation because it takes less space to write solution sets.

  2. 137. I can check inequalities by substituting 0 for the variable: When 0 belongs to the solution set, I should obtain a true statement, and when 0 does not belong to the solution set, I should obtain a false statement.

  3. 138. In an inequality such as 5x+4<8x5, I can avoid division by a negative number depending on which side I collect the variable terms and on which side I collect the constant terms.

  4. 139. I’ll win the contest if I can complete the crossword puzzle in 20 minutes plus or minus 5 minutes, so my winning time, x, is modeled by |x20|5.

In Exercises 140143, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 140. (, 1][4, )=[4, 1]

  2. 141. (, 3)(, 2)=(, 2)

  3. 142. The inequality 3x>6 is equivalent to 2>x.

  4. 143. All irrational numbers satisfy |x4|>0.

  5. 144. What’s wrong with this argument? Suppose x and y represent two real numbers, where x>y.

    2>1This is a true statement.2(yx)>1(yx)Multiply both sidesby yx.2y2x>yx Use the distributive property. y2x>xSubtract y from both sides.y>xAdd 2x to both sides.

    The final inequality, y>x, is impossible because we were initially given x>y.

  6. 145. Write an absolute value inequality for which the interval shown is the solution.

    Two number lines.

Group Exercise

  1. 146. Each group member should research one situation that provides two different pricing options. These can involve areas such as public transportation options (with or without electronic passes), cellphone plans, long-distance telephone plans, or anything of interest. Be sure to bring in all the details for each option. At a second group meeting, select the two pricing situations that are most interesting and relevant. Using each situation, write a word problem about selecting the better of the two options. The word problem should be one that can be solved using a linear inequality. The group should turn in the two problems and their solutions.

Preview Exercises

Exercises 147149 will help you prepare for the material covered in the first section of the next chapter.

  1. 147. If y=4x, find the value of y that corresponds to values of x for each integer starting with 3 and ending with 3.

  2. 148. If y=4x2, find the value of y that corresponds to values of x for each integer starting with 3 and ending with 3.

  3. 149. If y=|x+1|, find the value of y that corresponds to values of x for each integer starting with 4 and ending with 2.

P.9: Exercise Set

P.9 Exercise Set

Practice Exercises

In Exercises 114, express each interval in set-builder notation and graph the interval on a number line.

  1. 1. (1, 6]

  2. 2. (2, 4]

  3. 3. [5, 2)

  4. 4. [4, 3)

  5. 5. [3, 1]

  6. 6. [2, 5]

  7. 7. (2, )

  8. 8. (3, )

  9. 9. [3, )

  10. 10. [5, )

  11. 11. (, 3)

  12. 12. (, 2)

  13. 13. (, 5.5)

  14. 14. (, 3.5]

In Exercises 1526, use graphs to find each set.

  1. 15. (3, 0)[1, 2]

  2. 16. (4, 0)[2, 1]

  3. 17. (3, 0)[1, 2]

  4. 18. (4, 0)[2, 1]

  5. 19. (, 5)[1, 8)

  6. 20. (, 6)[2, 9)

  7. 21. (, 5)[1, 8)

  8. 22. (, 6)[2, 9)

  9. 23. [3, )(6, )

  10. 24. [2, )(4, )

  11. 25. [3, )(6, )

  12. 26. [2, )(4, )

In all exercises, use interval notation to express solution sets and graph each solution set on a number line.

In Exercises 2748, solve each linear inequality.

  1. 27. 5x+11<26

  2. 28. 2x+5<17

  3. 29. 3x713

  4. 30. 8x214

  5. 31. 9x36

  6. 32. 5x30

  7. 33. 8x113x13

  8. 34. 18x+4512x8

  9. 35. 4(x+1)+23x+6

  10. 36. 8x+3>3(2x+1)+x+5

  11. 37. 2x11<3(x+2)

  12. 38. 4(x+2)>3x+20

  13. 39. 1(x+3)42x

  14. 40. 5(3x)3x1

  15. 41. x432x2+1

  16. 42. 3x10+115x10

  17. 43. 1x2>4

  18. 44. 745 x<35

  19. 45. x46x29+518

  20. 46. 4x36+22x112

  21. 47. 3[3(x+5)+8x+7]+5[3(x6)2(3x5)]<2(4x+3)

  22. 48. 5[3(23x)2(5x)]6[5(x2)2(4x3)]<3x+19

In Exercises 4956, solve each compound inequality.

  1. 49. 6<x+3<8

  2. 50. 7<x+5<11

  3. 51. 3x2<1

  4. 52. 6<x41

  5. 53. 11<2x15

  6. 54. 34x3<19

  7. 55. 323 x5<1

  8. 56. 612 x4<3

In Exercises 5792, solve each absolute value inequality.

  1. 57. |x|<3

  2. 58. |x|<5

  3. 59. |x1|2

  4. 60. |x+3|4

  5. 61. |2x6|<8

  6. 62. |3x+5|<17

  7. 63. |2(x1)+4|8

  8. 64. |3(x1)+2|20

  9. 65. |2x+63|<2

  10. 66. |3(x1)4|<6

  11. 67. |x|>3

  12. 68. |x|>5

  13. 69. |x1|2

  14. 70. |x+3|4

  15. 71. |3x8|>7

  16. 72. |5x2|>13

  17. 73. |2x+24|2

  18. 74. |3x39|1

  19. 75. |323 x|>5

  20. 76. |334 x|>9

  21. 77. 3|x1|+28

  22. 78. 5|2x+1|39

  23. 79. 2|x4|4

  24. 80. 3|x+7|27

  25. 81. 4|1x|<16

  26. 82. 2|5x|<6

  27. 83. 3|2x1|

  28. 84. 9|4x+7|

  29. 85. 5>|4x|

  30. 86. 2>|11x|

  31. 87. 1<|23x|

  32. 88. 4<|2x|

  33. 89. 12<|2x+67|+37

  34. 90. 1<|x113|+73

  35. 91. 4+|3x3|9

  36. 92. |2x2|11

Practice PLUS

In Exercises 93100, use interval notation to represent all values of x satisfying the given conditions.

  1. 93. y1=x2+3, y2=x3+52, and y1y2.

  2. 94. y1=23 (6x9)+4, y2=5x+1, and y1>y2.

  3. 95. y=1(x+3)+2x and y is at least 4.

  4. 96. y=2x11+3(x+2) and y is at most 0.

  5. 97. y=|3x4|+2 and y<8.

  6. 98. y=|2x5|+1 and y>9.

  7. 99. y=7|x2+2| and y is at most 4.

  8. 100. y=8|5x+3| and y is at least 6.

  9. 101. When 3 times a number is subtracted from 4, the absolute value of the difference is at least 5. Use interval notation to express the set of all numbers that satisfy this condition.

  10. 102. When 4 times a number is subtracted from 5, the absolute value of the difference is at most 13. Use interval notation to express the set of all numbers that satisfy this condition.

Application Exercises

The graphs show that the three components of love, namely, passion, intimacy, and commitment, progress differently over time. Passion peaks early in a relationship and then declines. By contrast, intimacy and commitment build gradually. Use the graphs to solve Exercises 103110.

A line graph depicts the course of love over time.

Source: R. J. Sternberg. A Triangular Theory of Love, Psychological Review, 93, 119–135

  1. 103. Use interval notation to write an inequality that expresses for which years in a relationship intimacy is greater than commitment.

  2. 104. Use interval notation to write an inequality that expresses for which years in a relationship passion is greater than or equal to intimacy.

  3. 105. What is the relationship between passion and intimacy on the interval [5, 7)?

  4. 106. What is the relationship between intimacy and commitment on the interval [4, 7)?

  5. 107. What is the relationship between passion and commitment for {x|6<x<8}?

  6. 108. What is the relationship between passion and commitment for {x|7<x<9}?

  7. 109. What is the maximum level of intensity for passion? After how many years in a relationship does this occur?

  8. 110. After approximately how many years do levels of intensity for commitment exceed the maximum level of intensity for passion?

In more U.S. marriages, spouses have different faiths. The bar graph shows the percentage of households with an interfaith marriage in 1988 and 2012. Also shown is the percentage of households in which a person of faith is married to someone with no religion.

A bar chart shows percentage of U.S. households in which married couples do not share the same faith.

Source: General Social Survey, University of Chicago

The formula

I=14x+26

models the percentage of U.S. households with an interfaith marriage, I, x years after 1988. The formula

N=14x+6

models the percentage of U.S. households in which a person of faith is married to someone with no religion, N, x years after 1988.

Use these models to solve Exercises 111112.

  1. 111.

    1. In which years will more than 33% of U.S. households have an interfaith marriage?

    2. In which years will more than 14% of U.S. households have a person of faith married to someone with no religion?

    3. Based on your answers to parts (a) and (b), in which years will more than 33% of households have an interfaith marriage and more than 14% have a faith/no religion marriage?

    4. Based on your answers to parts (a) and (b), in which years will more than 33% of households have an interfaith marriage or more than 14% have a faith/no religion marriage?

  2. 112.

    1. In which years will more than 34% of U.S. households have an interfaith marriage?

    2. In which years will more than 15% of U.S. households have a person of faith married to someone with no religion?

    3. Based on your answers to parts (a) and (b), in which years will more than 34% of households have an interfaith marriage and more than 15% have a faith/no religion marriage?

    4. Based on your answers to parts (a) and (b), in which years will more than 34% of households have an interfaith marriage or more than 15% have a faith/no religion marriage?

  3. 113. The formula for converting Fahrenheit temperature, F, to Celsius temperature, C, is

    C=59 (F32).

    If Celsius temperature ranges from 15° to 35°, inclusive, what is the range for the Fahrenheit temperature? Use interval notation to express this range.

  4. 114. The formula for converting Celsius temperature, C, to Fahrenheit temperature, F, is

    F = 95 C+ 32.

    If Fahrenheit temperature ranges from 41° to 50°, inclusive, what is the range for Celsius temperature? Use interval notation to express this range.

  5. 115. If a coin is tossed 100 times, we would expect approximately 50 of the outcomes to be heads. It can be demonstrated that a coin is unfair if h, the number of outcomes that result in heads, satisfies |h505|1.645. Describe the number of outcomes that determine an unfair coin that is tossed 100 times.

In Exercises 116127, use the strategy for solving word problems, modeling the verbal conditions of the problem with a linear inequality.

  1. 116. You see a sign on a small moving truck that reads “Rent me for $20 a day.*” Oh, there’s an asterisk: There is an additional charge of $0.80 per mile. When you go online to reserve the truck, you are offered an unlimited mileage option for $60 a day. How many miles would you have to drive the truck in a day to make the unlimited mileage option the better deal?

  2. 117. A transponder for a toll bridge costs $27.50. With the transponder, the toll is $5 each time you cross the bridge. The only other option is toll-by-plate, for which the toll is $6.25 each time you cross the bridge with an additional administrative fee of $1.25 for each crossing. For how many bridge crossings is toll-by-plate the better option?

  3. 118. An electronic pass for a toll road costs $30. The toll is normally $5.00 but is reduced by 30% for people who have purchased the electronic pass. Determine the number of times the road must be used so that the electronic pass option is the better deal.

  4. 119. A city commission has proposed two tax bills. The first bill requires that a homeowner pay $1200 plus 0.5% of the assessed home value in taxes. The second bill requires taxes of $300 plus 0.9% of the assessed home value. What price range of home assessment would make the first bill a better deal?

  5. 120. A company designs and sells greeting cards. The weekly fixed cost is $10,000 and it costs $0.40 to create each card. The selling price is $2.00 per card. How many greeting cards must be designed and sold each week for the company to generate a profit?

  6. 121. A company manufactures and sells personalized stationery. The weekly fixed cost is $3000 and it costs $3.00 to produce each package of stationery. The selling price is $5.50 per package. How many packages of stationery must be produced and sold each week for the company to generate a profit?

  7. 122. An elevator at a construction site has a maximum capacity of 2800 pounds. If the elevator operator weighs 265 pounds and each cement bag weighs 65 pounds, how many bags of cement can be safely lifted on the elevator in one trip?

  8. 123. An elevator at a construction site has a maximum capacity of 3000 pounds. If the elevator operator weighs 245 pounds and each cement bag weighs 95 pounds, how many bags of cement can be safely lifted on the elevator in one trip?

  9. 124. To earn an A in a course, you must have a final average of at least 90%. On the first four examinations, you have grades of 86%, 88%, 92%, and 84%. If the final examination counts as two grades, what must you get on the final to earn an A in the course?

  10. 125. On two examinations, you have grades of 86 and 88. There is an optional final examination, which counts as one grade. You decide to take the final in order to get a course grade of A, meaning a final average of at least 90.

    1. What must you get on the final to earn an A in the course?

    2. By taking the final, if you do poorly, you might risk the B that you have in the course based on the first two exam grades. If your final average is less than 80, you will lose your B in the course. Describe the grades on the final that will cause this to happen.

  11. 126. Parts for an automobile repair cost $254. The mechanic charges $65 per hour. If you receive an estimate for at least $351.50 and at most $481.50 for fixing the car, what is the time interval that the mechanic will be working on the job?

  12. 127. The toll to a bridge is $3.00. A three-month pass costs $7.50 and reduces the toll to $0.50. A six-month pass costs $30 and permits crossing the bridge for no additional fee. How many crossings per three-month period does it take for the three-month pass to be the best deal?

Explaining the Concepts

  1. 128. When graphing the solutions of an inequality, what does a parenthesis signify? What does a square bracket signify?

  2. 129. Describe ways in which solving a linear inequality is similar to solving a linear equation.

  3. 130. Describe ways in which solving a linear inequality is different from solving a linear equation.

  4. 131. What is a compound inequality and how is it solved?

  5. 132. Describe how to solve an absolute value inequality involving the symbol <. Give an example.

  6. 133. Describe how to solve an absolute value inequality involving the symbol >. Give an example.

  7. 134. Explain why |x|<4 has no solution.

  8. 135. Describe the solution set of |x|>4.

Critical Thinking Exercises

Make Sense? In Exercises 136139, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 136. I prefer interval notation over set-builder notation because it takes less space to write solution sets.

  2. 137. I can check inequalities by substituting 0 for the variable: When 0 belongs to the solution set, I should obtain a true statement, and when 0 does not belong to the solution set, I should obtain a false statement.

  3. 138. In an inequality such as 5x+4<8x5, I can avoid division by a negative number depending on which side I collect the variable terms and on which side I collect the constant terms.

  4. 139. I’ll win the contest if I can complete the crossword puzzle in 20 minutes plus or minus 5 minutes, so my winning time, x, is modeled by |x20|5.

In Exercises 140143, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 140. (, 1][4, )=[4, 1]

  2. 141. (, 3)(, 2)=(, 2)

  3. 142. The inequality 3x>6 is equivalent to 2>x.

  4. 143. All irrational numbers satisfy |x4|>0.

  5. 144. What’s wrong with this argument? Suppose x and y represent two real numbers, where x>y.

    2>1This is a true statement.2(yx)>1(yx)Multiply both sidesby yx.2y2x>yx Use the distributive property. y2x>xSubtract y from both sides.y>xAdd 2x to both sides.

    The final inequality, y>x, is impossible because we were initially given x>y.

  6. 145. Write an absolute value inequality for which the interval shown is the solution.

    Two number lines.

Group Exercise

  1. 146. Each group member should research one situation that provides two different pricing options. These can involve areas such as public transportation options (with or without electronic passes), cellphone plans, long-distance telephone plans, or anything of interest. Be sure to bring in all the details for each option. At a second group meeting, select the two pricing situations that are most interesting and relevant. Using each situation, write a word problem about selecting the better of the two options. The word problem should be one that can be solved using a linear inequality. The group should turn in the two problems and their solutions.

Preview Exercises

Exercises 147149 will help you prepare for the material covered in the first section of the next chapter.

  1. 147. If y=4x, find the value of y that corresponds to values of x for each integer starting with 3 and ending with 3.

  2. 148. If y=4x2, find the value of y that corresponds to values of x for each integer starting with 3 and ending with 3.

  3. 149. If y=|x+1|, find the value of y that corresponds to values of x for each integer starting with 4 and ending with 2.

P.9: Exercise Set

P.9 Exercise Set

Practice Exercises

In Exercises 114, express each interval in set-builder notation and graph the interval on a number line.

  1. 1. (1, 6]

  2. 2. (2, 4]

  3. 3. [5, 2)

  4. 4. [4, 3)

  5. 5. [3, 1]

  6. 6. [2, 5]

  7. 7. (2, )

  8. 8. (3, )

  9. 9. [3, )

  10. 10. [5, )

  11. 11. (, 3)

  12. 12. (, 2)

  13. 13. (, 5.5)

  14. 14. (, 3.5]

In Exercises 1526, use graphs to find each set.

  1. 15. (3, 0)[1, 2]

  2. 16. (4, 0)[2, 1]

  3. 17. (3, 0)[1, 2]

  4. 18. (4, 0)[2, 1]

  5. 19. (, 5)[1, 8)

  6. 20. (, 6)[2, 9)

  7. 21. (, 5)[1, 8)

  8. 22. (, 6)[2, 9)

  9. 23. [3, )(6, )

  10. 24. [2, )(4, )

  11. 25. [3, )(6, )

  12. 26. [2, )(4, )

In all exercises, use interval notation to express solution sets and graph each solution set on a number line.

In Exercises 2748, solve each linear inequality.

  1. 27. 5x+11<26

  2. 28. 2x+5<17

  3. 29. 3x713

  4. 30. 8x214

  5. 31. 9x36

  6. 32. 5x30

  7. 33. 8x113x13

  8. 34. 18x+4512x8

  9. 35. 4(x+1)+23x+6

  10. 36. 8x+3>3(2x+1)+x+5

  11. 37. 2x11<3(x+2)

  12. 38. 4(x+2)>3x+20

  13. 39. 1(x+3)42x

  14. 40. 5(3x)3x1

  15. 41. x432x2+1

  16. 42. 3x10+115x10

  17. 43. 1x2>4

  18. 44. 745 x<35

  19. 45. x46x29+518

  20. 46. 4x36+22x112

  21. 47. 3[3(x+5)+8x+7]+5[3(x6)2(3x5)]<2(4x+3)

  22. 48. 5[3(23x)2(5x)]6[5(x2)2(4x3)]<3x+19

In Exercises 4956, solve each compound inequality.

  1. 49. 6<x+3<8

  2. 50. 7<x+5<11

  3. 51. 3x2<1

  4. 52. 6<x41

  5. 53. 11<2x15

  6. 54. 34x3<19

  7. 55. 323 x5<1

  8. 56. 612 x4<3

In Exercises 5792, solve each absolute value inequality.

  1. 57. |x|<3

  2. 58. |x|<5

  3. 59. |x1|2

  4. 60. |x+3|4

  5. 61. |2x6|<8

  6. 62. |3x+5|<17

  7. 63. |2(x1)+4|8

  8. 64. |3(x1)+2|20

  9. 65. |2x+63|<2

  10. 66. |3(x1)4|<6

  11. 67. |x|>3

  12. 68. |x|>5

  13. 69. |x1|2

  14. 70. |x+3|4

  15. 71. |3x8|>7

  16. 72. |5x2|>13

  17. 73. |2x+24|2

  18. 74. |3x39|1

  19. 75. |323 x|>5

  20. 76. |334 x|>9

  21. 77. 3|x1|+28

  22. 78. 5|2x+1|39

  23. 79. 2|x4|4

  24. 80. 3|x+7|27

  25. 81. 4|1x|<16

  26. 82. 2|5x|<6

  27. 83. 3|2x1|

  28. 84. 9|4x+7|

  29. 85. 5>|4x|

  30. 86. 2>|11x|

  31. 87. 1<|23x|

  32. 88. 4<|2x|

  33. 89. 12<|2x+67|+37

  34. 90. 1<|x113|+73

  35. 91. 4+|3x3|9

  36. 92. |2x2|11

Practice PLUS

In Exercises 93100, use interval notation to represent all values of x satisfying the given conditions.

  1. 93. y1=x2+3, y2=x3+52, and y1y2.

  2. 94. y1=23 (6x9)+4, y2=5x+1, and y1>y2.

  3. 95. y=1(x+3)+2x and y is at least 4.

  4. 96. y=2x11+3(x+2) and y is at most 0.

  5. 97. y=|3x4|+2 and y<8.

  6. 98. y=|2x5|+1 and y>9.

  7. 99. y=7|x2+2| and y is at most 4.

  8. 100. y=8|5x+3| and y is at least 6.

  9. 101. When 3 times a number is subtracted from 4, the absolute value of the difference is at least 5. Use interval notation to express the set of all numbers that satisfy this condition.

  10. 102. When 4 times a number is subtracted from 5, the absolute value of the difference is at most 13. Use interval notation to express the set of all numbers that satisfy this condition.

Application Exercises

The graphs show that the three components of love, namely, passion, intimacy, and commitment, progress differently over time. Passion peaks early in a relationship and then declines. By contrast, intimacy and commitment build gradually. Use the graphs to solve Exercises 103110.

A line graph depicts the course of love over time.

Source: R. J. Sternberg. A Triangular Theory of Love, Psychological Review, 93, 119–135

  1. 103. Use interval notation to write an inequality that expresses for which years in a relationship intimacy is greater than commitment.

  2. 104. Use interval notation to write an inequality that expresses for which years in a relationship passion is greater than or equal to intimacy.

  3. 105. What is the relationship between passion and intimacy on the interval [5, 7)?

  4. 106. What is the relationship between intimacy and commitment on the interval [4, 7)?

  5. 107. What is the relationship between passion and commitment for {x|6<x<8}?

  6. 108. What is the relationship between passion and commitment for {x|7<x<9}?

  7. 109. What is the maximum level of intensity for passion? After how many years in a relationship does this occur?

  8. 110. After approximately how many years do levels of intensity for commitment exceed the maximum level of intensity for passion?

In more U.S. marriages, spouses have different faiths. The bar graph shows the percentage of households with an interfaith marriage in 1988 and 2012. Also shown is the percentage of households in which a person of faith is married to someone with no religion.

A bar chart shows percentage of U.S. households in which married couples do not share the same faith.

Source: General Social Survey, University of Chicago

The formula

I=14x+26

models the percentage of U.S. households with an interfaith marriage, I, x years after 1988. The formula

N=14x+6

models the percentage of U.S. households in which a person of faith is married to someone with no religion, N, x years after 1988.

Use these models to solve Exercises 111112.

  1. 111.

    1. In which years will more than 33% of U.S. households have an interfaith marriage?

    2. In which years will more than 14% of U.S. households have a person of faith married to someone with no religion?

    3. Based on your answers to parts (a) and (b), in which years will more than 33% of households have an interfaith marriage and more than 14% have a faith/no religion marriage?

    4. Based on your answers to parts (a) and (b), in which years will more than 33% of households have an interfaith marriage or more than 14% have a faith/no religion marriage?

  2. 112.

    1. In which years will more than 34% of U.S. households have an interfaith marriage?

    2. In which years will more than 15% of U.S. households have a person of faith married to someone with no religion?

    3. Based on your answers to parts (a) and (b), in which years will more than 34% of households have an interfaith marriage and more than 15% have a faith/no religion marriage?

    4. Based on your answers to parts (a) and (b), in which years will more than 34% of households have an interfaith marriage or more than 15% have a faith/no religion marriage?

  3. 113. The formula for converting Fahrenheit temperature, F, to Celsius temperature, C, is

    C=59 (F32).

    If Celsius temperature ranges from 15° to 35°, inclusive, what is the range for the Fahrenheit temperature? Use interval notation to express this range.

  4. 114. The formula for converting Celsius temperature, C, to Fahrenheit temperature, F, is

    F = 95 C+ 32.

    If Fahrenheit temperature ranges from 41° to 50°, inclusive, what is the range for Celsius temperature? Use interval notation to express this range.

  5. 115. If a coin is tossed 100 times, we would expect approximately 50 of the outcomes to be heads. It can be demonstrated that a coin is unfair if h, the number of outcomes that result in heads, satisfies |h505|1.645. Describe the number of outcomes that determine an unfair coin that is tossed 100 times.

In Exercises 116127, use the strategy for solving word problems, modeling the verbal conditions of the problem with a linear inequality.

  1. 116. You see a sign on a small moving truck that reads “Rent me for $20 a day.*” Oh, there’s an asterisk: There is an additional charge of $0.80 per mile. When you go online to reserve the truck, you are offered an unlimited mileage option for $60 a day. How many miles would you have to drive the truck in a day to make the unlimited mileage option the better deal?

  2. 117. A transponder for a toll bridge costs $27.50. With the transponder, the toll is $5 each time you cross the bridge. The only other option is toll-by-plate, for which the toll is $6.25 each time you cross the bridge with an additional administrative fee of $1.25 for each crossing. For how many bridge crossings is toll-by-plate the better option?

  3. 118. An electronic pass for a toll road costs $30. The toll is normally $5.00 but is reduced by 30% for people who have purchased the electronic pass. Determine the number of times the road must be used so that the electronic pass option is the better deal.

  4. 119. A city commission has proposed two tax bills. The first bill requires that a homeowner pay $1200 plus 0.5% of the assessed home value in taxes. The second bill requires taxes of $300 plus 0.9% of the assessed home value. What price range of home assessment would make the first bill a better deal?

  5. 120. A company designs and sells greeting cards. The weekly fixed cost is $10,000 and it costs $0.40 to create each card. The selling price is $2.00 per card. How many greeting cards must be designed and sold each week for the company to generate a profit?

  6. 121. A company manufactures and sells personalized stationery. The weekly fixed cost is $3000 and it costs $3.00 to produce each package of stationery. The selling price is $5.50 per package. How many packages of stationery must be produced and sold each week for the company to generate a profit?

  7. 122. An elevator at a construction site has a maximum capacity of 2800 pounds. If the elevator operator weighs 265 pounds and each cement bag weighs 65 pounds, how many bags of cement can be safely lifted on the elevator in one trip?

  8. 123. An elevator at a construction site has a maximum capacity of 3000 pounds. If the elevator operator weighs 245 pounds and each cement bag weighs 95 pounds, how many bags of cement can be safely lifted on the elevator in one trip?

  9. 124. To earn an A in a course, you must have a final average of at least 90%. On the first four examinations, you have grades of 86%, 88%, 92%, and 84%. If the final examination counts as two grades, what must you get on the final to earn an A in the course?

  10. 125. On two examinations, you have grades of 86 and 88. There is an optional final examination, which counts as one grade. You decide to take the final in order to get a course grade of A, meaning a final average of at least 90.

    1. What must you get on the final to earn an A in the course?

    2. By taking the final, if you do poorly, you might risk the B that you have in the course based on the first two exam grades. If your final average is less than 80, you will lose your B in the course. Describe the grades on the final that will cause this to happen.

  11. 126. Parts for an automobile repair cost $254. The mechanic charges $65 per hour. If you receive an estimate for at least $351.50 and at most $481.50 for fixing the car, what is the time interval that the mechanic will be working on the job?

  12. 127. The toll to a bridge is $3.00. A three-month pass costs $7.50 and reduces the toll to $0.50. A six-month pass costs $30 and permits crossing the bridge for no additional fee. How many crossings per three-month period does it take for the three-month pass to be the best deal?

Explaining the Concepts

  1. 128. When graphing the solutions of an inequality, what does a parenthesis signify? What does a square bracket signify?

  2. 129. Describe ways in which solving a linear inequality is similar to solving a linear equation.

  3. 130. Describe ways in which solving a linear inequality is different from solving a linear equation.

  4. 131. What is a compound inequality and how is it solved?

  5. 132. Describe how to solve an absolute value inequality involving the symbol <. Give an example.

  6. 133. Describe how to solve an absolute value inequality involving the symbol >. Give an example.

  7. 134. Explain why |x|<4 has no solution.

  8. 135. Describe the solution set of |x|>4.

Critical Thinking Exercises

Make Sense? In Exercises 136139, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 136. I prefer interval notation over set-builder notation because it takes less space to write solution sets.

  2. 137. I can check inequalities by substituting 0 for the variable: When 0 belongs to the solution set, I should obtain a true statement, and when 0 does not belong to the solution set, I should obtain a false statement.

  3. 138. In an inequality such as 5x+4<8x5, I can avoid division by a negative number depending on which side I collect the variable terms and on which side I collect the constant terms.

  4. 139. I’ll win the contest if I can complete the crossword puzzle in 20 minutes plus or minus 5 minutes, so my winning time, x, is modeled by |x20|5.

In Exercises 140143, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 140. (, 1][4, )=[4, 1]

  2. 141. (, 3)(, 2)=(, 2)

  3. 142. The inequality 3x>6 is equivalent to 2>x.

  4. 143. All irrational numbers satisfy |x4|>0.

  5. 144. What’s wrong with this argument? Suppose x and y represent two real numbers, where x>y.

    2>1This is a true statement.2(yx)>1(yx)Multiply both sidesby yx.2y2x>yx Use the distributive property. y2x>xSubtract y from both sides.y>xAdd 2x to both sides.

    The final inequality, y>x, is impossible because we were initially given x>y.

  6. 145. Write an absolute value inequality for which the interval shown is the solution.

    Two number lines.

Group Exercise

  1. 146. Each group member should research one situation that provides two different pricing options. These can involve areas such as public transportation options (with or without electronic passes), cellphone plans, long-distance telephone plans, or anything of interest. Be sure to bring in all the details for each option. At a second group meeting, select the two pricing situations that are most interesting and relevant. Using each situation, write a word problem about selecting the better of the two options. The word problem should be one that can be solved using a linear inequality. The group should turn in the two problems and their solutions.

Preview Exercises

Exercises 147149 will help you prepare for the material covered in the first section of the next chapter.

  1. 147. If y=4x, find the value of y that corresponds to values of x for each integer starting with 3 and ending with 3.

  2. 148. If y=4x2, find the value of y that corresponds to values of x for each integer starting with 3 and ending with 3.

  3. 149. If y=|x+1|, find the value of y that corresponds to values of x for each integer starting with 4 and ending with 2.

P.9: Exercise Set

P.9 Exercise Set

Practice Exercises

In Exercises 114, express each interval in set-builder notation and graph the interval on a number line.

  1. 1. (1, 6]

  2. 2. (2, 4]

  3. 3. [5, 2)

  4. 4. [4, 3)

  5. 5. [3, 1]

  6. 6. [2, 5]

  7. 7. (2, )

  8. 8. (3, )

  9. 9. [3, )

  10. 10. [5, )

  11. 11. (, 3)

  12. 12. (, 2)

  13. 13. (, 5.5)

  14. 14. (, 3.5]

In Exercises 1526, use graphs to find each set.

  1. 15. (3, 0)[1, 2]

  2. 16. (4, 0)[2, 1]

  3. 17. (3, 0)[1, 2]

  4. 18. (4, 0)[2, 1]

  5. 19. (, 5)[1, 8)

  6. 20. (, 6)[2, 9)

  7. 21. (, 5)[1, 8)

  8. 22. (, 6)[2, 9)

  9. 23. [3, )(6, )

  10. 24. [2, )(4, )

  11. 25. [3, )(6, )

  12. 26. [2, )(4, )

In all exercises, use interval notation to express solution sets and graph each solution set on a number line.

In Exercises 2748, solve each linear inequality.

  1. 27. 5x+11<26

  2. 28. 2x+5<17

  3. 29. 3x713

  4. 30. 8x214

  5. 31. 9x36

  6. 32. 5x30

  7. 33. 8x113x13

  8. 34. 18x+4512x8

  9. 35. 4(x+1)+23x+6

  10. 36. 8x+3>3(2x+1)+x+5

  11. 37. 2x11<3(x+2)

  12. 38. 4(x+2)>3x+20

  13. 39. 1(x+3)42x

  14. 40. 5(3x)3x1

  15. 41. x432x2+1

  16. 42. 3x10+115x10

  17. 43. 1x2>4

  18. 44. 745 x<35

  19. 45. x46x29+518

  20. 46. 4x36+22x112

  21. 47. 3[3(x+5)+8x+7]+5[3(x6)2(3x5)]<2(4x+3)

  22. 48. 5[3(23x)2(5x)]6[5(x2)2(4x3)]<3x+19

In Exercises 4956, solve each compound inequality.

  1. 49. 6<x+3<8

  2. 50. 7<x+5<11

  3. 51. 3x2<1

  4. 52. 6<x41

  5. 53. 11<2x15

  6. 54. 34x3<19

  7. 55. 323 x5<1

  8. 56. 612 x4<3

In Exercises 5792, solve each absolute value inequality.

  1. 57. |x|<3

  2. 58. |x|<5

  3. 59. |x1|2

  4. 60. |x+3|4

  5. 61. |2x6|<8

  6. 62. |3x+5|<17

  7. 63. |2(x1)+4|8

  8. 64. |3(x1)+2|20

  9. 65. |2x+63|<2

  10. 66. |3(x1)4|<6

  11. 67. |x|>3

  12. 68. |x|>5

  13. 69. |x1|2

  14. 70. |x+3|4

  15. 71. |3x8|>7

  16. 72. |5x2|>13

  17. 73. |2x+24|2

  18. 74. |3x39|1

  19. 75. |323 x|>5

  20. 76. |334 x|>9

  21. 77. 3|x1|+28

  22. 78. 5|2x+1|39

  23. 79. 2|x4|4

  24. 80. 3|x+7|27

  25. 81. 4|1x|<16

  26. 82. 2|5x|<6

  27. 83. 3|2x1|

  28. 84. 9|4x+7|

  29. 85. 5>|4x|

  30. 86. 2>|11x|

  31. 87. 1<|23x|

  32. 88. 4<|2x|

  33. 89. 12<|2x+67|+37

  34. 90. 1<|x113|+73

  35. 91. 4+|3x3|9

  36. 92. |2x2|11

Practice PLUS

In Exercises 93100, use interval notation to represent all values of x satisfying the given conditions.

  1. 93. y1=x2+3, y2=x3+52, and y1y2.

  2. 94. y1=23 (6x9)+4, y2=5x+1, and y1>y2.

  3. 95. y=1(x+3)+2x and y is at least 4.

  4. 96. y=2x11+3(x+2) and y is at most 0.

  5. 97. y=|3x4|+2 and y<8.

  6. 98. y=|2x5|+1 and y>9.

  7. 99. y=7|x2+2| and y is at most 4.

  8. 100. y=8|5x+3| and y is at least 6.

  9. 101. When 3 times a number is subtracted from 4, the absolute value of the difference is at least 5. Use interval notation to express the set of all numbers that satisfy this condition.

  10. 102. When 4 times a number is subtracted from 5, the absolute value of the difference is at most 13. Use interval notation to express the set of all numbers that satisfy this condition.

Application Exercises

The graphs show that the three components of love, namely, passion, intimacy, and commitment, progress differently over time. Passion peaks early in a relationship and then declines. By contrast, intimacy and commitment build gradually. Use the graphs to solve Exercises 103110.

A line graph depicts the course of love over time.

Source: R. J. Sternberg. A Triangular Theory of Love, Psychological Review, 93, 119–135

  1. 103. Use interval notation to write an inequality that expresses for which years in a relationship intimacy is greater than commitment.

  2. 104. Use interval notation to write an inequality that expresses for which years in a relationship passion is greater than or equal to intimacy.

  3. 105. What is the relationship between passion and intimacy on the interval [5, 7)?

  4. 106. What is the relationship between intimacy and commitment on the interval [4, 7)?

  5. 107. What is the relationship between passion and commitment for {x|6<x<8}?

  6. 108. What is the relationship between passion and commitment for {x|7<x<9}?

  7. 109. What is the maximum level of intensity for passion? After how many years in a relationship does this occur?

  8. 110. After approximately how many years do levels of intensity for commitment exceed the maximum level of intensity for passion?

In more U.S. marriages, spouses have different faiths. The bar graph shows the percentage of households with an interfaith marriage in 1988 and 2012. Also shown is the percentage of households in which a person of faith is married to someone with no religion.

A bar chart shows percentage of U.S. households in which married couples do not share the same faith.

Source: General Social Survey, University of Chicago

The formula

I=14x+26

models the percentage of U.S. households with an interfaith marriage, I, x years after 1988. The formula

N=14x+6

models the percentage of U.S. households in which a person of faith is married to someone with no religion, N, x years after 1988.

Use these models to solve Exercises 111112.

  1. 111.

    1. In which years will more than 33% of U.S. households have an interfaith marriage?

    2. In which years will more than 14% of U.S. households have a person of faith married to someone with no religion?

    3. Based on your answers to parts (a) and (b), in which years will more than 33% of households have an interfaith marriage and more than 14% have a faith/no religion marriage?

    4. Based on your answers to parts (a) and (b), in which years will more than 33% of households have an interfaith marriage or more than 14% have a faith/no religion marriage?

  2. 112.

    1. In which years will more than 34% of U.S. households have an interfaith marriage?

    2. In which years will more than 15% of U.S. households have a person of faith married to someone with no religion?

    3. Based on your answers to parts (a) and (b), in which years will more than 34% of households have an interfaith marriage and more than 15% have a faith/no religion marriage?

    4. Based on your answers to parts (a) and (b), in which years will more than 34% of households have an interfaith marriage or more than 15% have a faith/no religion marriage?

  3. 113. The formula for converting Fahrenheit temperature, F, to Celsius temperature, C, is

    C=59 (F32).

    If Celsius temperature ranges from 15° to 35°, inclusive, what is the range for the Fahrenheit temperature? Use interval notation to express this range.

  4. 114. The formula for converting Celsius temperature, C, to Fahrenheit temperature, F, is

    F = 95 C+ 32.

    If Fahrenheit temperature ranges from 41° to 50°, inclusive, what is the range for Celsius temperature? Use interval notation to express this range.

  5. 115. If a coin is tossed 100 times, we would expect approximately 50 of the outcomes to be heads. It can be demonstrated that a coin is unfair if h, the number of outcomes that result in heads, satisfies |h505|1.645. Describe the number of outcomes that determine an unfair coin that is tossed 100 times.

In Exercises 116127, use the strategy for solving word problems, modeling the verbal conditions of the problem with a linear inequality.

  1. 116. You see a sign on a small moving truck that reads “Rent me for $20 a day.*” Oh, there’s an asterisk: There is an additional charge of $0.80 per mile. When you go online to reserve the truck, you are offered an unlimited mileage option for $60 a day. How many miles would you have to drive the truck in a day to make the unlimited mileage option the better deal?

  2. 117. A transponder for a toll bridge costs $27.50. With the transponder, the toll is $5 each time you cross the bridge. The only other option is toll-by-plate, for which the toll is $6.25 each time you cross the bridge with an additional administrative fee of $1.25 for each crossing. For how many bridge crossings is toll-by-plate the better option?

  3. 118. An electronic pass for a toll road costs $30. The toll is normally $5.00 but is reduced by 30% for people who have purchased the electronic pass. Determine the number of times the road must be used so that the electronic pass option is the better deal.

  4. 119. A city commission has proposed two tax bills. The first bill requires that a homeowner pay $1200 plus 0.5% of the assessed home value in taxes. The second bill requires taxes of $300 plus 0.9% of the assessed home value. What price range of home assessment would make the first bill a better deal?

  5. 120. A company designs and sells greeting cards. The weekly fixed cost is $10,000 and it costs $0.40 to create each card. The selling price is $2.00 per card. How many greeting cards must be designed and sold each week for the company to generate a profit?

  6. 121. A company manufactures and sells personalized stationery. The weekly fixed cost is $3000 and it costs $3.00 to produce each package of stationery. The selling price is $5.50 per package. How many packages of stationery must be produced and sold each week for the company to generate a profit?

  7. 122. An elevator at a construction site has a maximum capacity of 2800 pounds. If the elevator operator weighs 265 pounds and each cement bag weighs 65 pounds, how many bags of cement can be safely lifted on the elevator in one trip?

  8. 123. An elevator at a construction site has a maximum capacity of 3000 pounds. If the elevator operator weighs 245 pounds and each cement bag weighs 95 pounds, how many bags of cement can be safely lifted on the elevator in one trip?

  9. 124. To earn an A in a course, you must have a final average of at least 90%. On the first four examinations, you have grades of 86%, 88%, 92%, and 84%. If the final examination counts as two grades, what must you get on the final to earn an A in the course?

  10. 125. On two examinations, you have grades of 86 and 88. There is an optional final examination, which counts as one grade. You decide to take the final in order to get a course grade of A, meaning a final average of at least 90.

    1. What must you get on the final to earn an A in the course?

    2. By taking the final, if you do poorly, you might risk the B that you have in the course based on the first two exam grades. If your final average is less than 80, you will lose your B in the course. Describe the grades on the final that will cause this to happen.

  11. 126. Parts for an automobile repair cost $254. The mechanic charges $65 per hour. If you receive an estimate for at least $351.50 and at most $481.50 for fixing the car, what is the time interval that the mechanic will be working on the job?

  12. 127. The toll to a bridge is $3.00. A three-month pass costs $7.50 and reduces the toll to $0.50. A six-month pass costs $30 and permits crossing the bridge for no additional fee. How many crossings per three-month period does it take for the three-month pass to be the best deal?

Explaining the Concepts

  1. 128. When graphing the solutions of an inequality, what does a parenthesis signify? What does a square bracket signify?

  2. 129. Describe ways in which solving a linear inequality is similar to solving a linear equation.

  3. 130. Describe ways in which solving a linear inequality is different from solving a linear equation.

  4. 131. What is a compound inequality and how is it solved?

  5. 132. Describe how to solve an absolute value inequality involving the symbol <. Give an example.

  6. 133. Describe how to solve an absolute value inequality involving the symbol >. Give an example.

  7. 134. Explain why |x|<4 has no solution.

  8. 135. Describe the solution set of |x|>4.

Critical Thinking Exercises

Make Sense? In Exercises 136139, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 136. I prefer interval notation over set-builder notation because it takes less space to write solution sets.

  2. 137. I can check inequalities by substituting 0 for the variable: When 0 belongs to the solution set, I should obtain a true statement, and when 0 does not belong to the solution set, I should obtain a false statement.

  3. 138. In an inequality such as 5x+4<8x5, I can avoid division by a negative number depending on which side I collect the variable terms and on which side I collect the constant terms.

  4. 139. I’ll win the contest if I can complete the crossword puzzle in 20 minutes plus or minus 5 minutes, so my winning time, x, is modeled by |x20|5.

In Exercises 140143, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 140. (, 1][4, )=[4, 1]

  2. 141. (, 3)(, 2)=(, 2)

  3. 142. The inequality 3x>6 is equivalent to 2>x.

  4. 143. All irrational numbers satisfy |x4|>0.

  5. 144. What’s wrong with this argument? Suppose x and y represent two real numbers, where x>y.

    2>1This is a true statement.2(yx)>1(yx)Multiply both sidesby yx.2y2x>yx Use the distributive property. y2x>xSubtract y from both sides.y>xAdd 2x to both sides.

    The final inequality, y>x, is impossible because we were initially given x>y.

  6. 145. Write an absolute value inequality for which the interval shown is the solution.

    Two number lines.

Group Exercise

  1. 146. Each group member should research one situation that provides two different pricing options. These can involve areas such as public transportation options (with or without electronic passes), cellphone plans, long-distance telephone plans, or anything of interest. Be sure to bring in all the details for each option. At a second group meeting, select the two pricing situations that are most interesting and relevant. Using each situation, write a word problem about selecting the better of the two options. The word problem should be one that can be solved using a linear inequality. The group should turn in the two problems and their solutions.

Preview Exercises

Exercises 147149 will help you prepare for the material covered in the first section of the next chapter.

  1. 147. If y=4x, find the value of y that corresponds to values of x for each integer starting with 3 and ending with 3.

  2. 148. If y=4x2, find the value of y that corresponds to values of x for each integer starting with 3 and ending with 3.

  3. 149. If y=|x+1|, find the value of y that corresponds to values of x for each integer starting with 4 and ending with 2.

Chapter P: Review Exercises

Chapter P Review Exercises

You can use these review exercises, like the review exercises at the end of each chapter, to test your understanding of the chapter’s topics. However, you can also use these exercises as a prerequisite test to check your mastery of the fundamental algebra skills needed in this book.

P.1

In Exercises 12, evaluate each algebraic expression for the given value or values of the variable(s).

  1. 1. 3+6(x2)3 for x=4

  2. 2. x25(xy) for x=6 and y=2

  3. 3. You are riding along an expressway traveling x miles per hour. The formula

    S=0.015x2+x+10

    models the recommended safe distance, S, in feet, between your car and other cars on the expressway. What is the recommended safe distance when your speed is 60 miles per hour?

In Exercises 47, let A={a, b, c}, B={a, c, d, e}, and C={a, d, f, g}. Find the indicated set.

  1. 4. AB

  2. 5. AB

  3. 6. AC

  4. 7. CA

  5. 8. Consider the set:

    {17, 913, 0, 0.75, 2, π, 81}.

    List all numbers from the set that are a. natural numbers, b. whole numbers, c. integers, d. rational numbers, e. irrational numbers, f. real numbers.

In Exercises 911, rewrite each expression without absolute value bars.

  1. 9. |103|

  2. 10. |21|

  3. 11. |317|

  4. 12. Express the distance between the numbers 17 and 4 using absolute value. Then evaluate the absolute value.

In Exercises 1318, state the name of the property illustrated.

  1. 13. 3+17=17+3

  2. 14. (63)9=6(39)

  3. 15. 3(5+3)=15+3

  4. 16. (69)2=2(69)

  5. 17. 3(5+3)=(5+3)3

  6. 18. (37)+(47)=(47)+(37)

In Exercises 1922, simplify each algebraic expression.

  1. 19. 5(2x3)+7x

  2. 20. 15(5x)+[(3y)+(3y)](x)

  3. 21. 3(4y5)(7y+2)

  4. 22. 82[3(5x1)]

  5. 23. The diversity index, from 0 (no diversity) to 100, measures the chance that two randomly selected people are a different race or ethnicity. The diversity index in the United States varies widely from region to region, from as high as 81 in Hawaii to as low as 11 in Vermont. The bar graph shows the national diversity index for the United States for five years in the period from 1980 through 2020.

    A bar graph shows the national diversity index for US for five years.

    Source: USA Today

    The data in the graph can be modeled by the formula

    D=0.005x2+0.55x+34,

    where D is the national diversity index in the United States x years after 1980. According to the formula, what was the U.S. diversity index in 2010? How does this compare with the index displayed by the bar graph?

P.2

Evaluate each exponential expression in Exercises 2427.

  1. 24. (3)3(2)2

  2. 25. 24+41

  3. 26. 535

  4. 27. 3336

Simplify each exponential expression in Exercises 2831.

  1. 28. (2x4 y3)3

  2. 29. (5x3 y2)(2x11 y2)

  3. 30. (2x3)4

  4. 31. 7x5 y628x15 y2

In Exercises 3233, write each number in decimal notation.

  1. 32. 3.74×104

  2. 33. 7.45×105

In Exercises 3435, write each number in scientific notation.

  1. 34. 3,590,000

  2. 35. 0.00725

In Exercises 3637, perform the indicated operation and write the answer in decimal notation.

  1. 36. (3×103)(1.3×102)

  2. 37. 6.9×1033×105

  3. 38. The average salary of a professional baseball player is $4.1 million. (Source: Major League Baseball Player Association) Express this number in scientific notation.

  4. 39. The average salary of a nurse is $73,000. (Source: U.S. Department of Labor) Express this number in scientific notation.

  5. 40. Use your scientific notation answers from Exercises 38 and 39 to answer this question.

    How many times greater is the average salary of a professional baseball player than the average salary of a nurse?

P.3

Use the product rule to simplify the expressions in Exercises 4144. In Exercises 4344, assume that variables represent nonnegative real numbers.

  1. 41. 300

  2. 42. 12x2

  3. 43. 10x2x

  4. 44. r3

Use the quotient rule to simplify the expressions in Exercises 4546.

  1. 45. 1214

  2. 46. 96x32x (Assume that x>0.)

In Exercises 4749, add or subtract terms whenever possible.

  1. 47. 75+135

  2. 48. 250+38

  3. 49. 472248

In Exercises 5053, rationalize the denominator.

  1. 50. 305

  2. 51. 23

  3. 52. 56+3

  4. 53. 1475

Evaluate each expression in Exercises 5457 or indicate that the root is not a real number.

  1. 54. 1253

  2. 55. 325

  3. 56. 1254

  4. 57. (5)44

Simplify the radical expressions in Exercises 5862.

  1. 58. 813

  2. 59. y53

  3. 60. 84104

  4. 61. 4163+523

  5. 62. 32x5416x4 (Assume that x>0.)

In Exercises 6368, evaluate each expression.

  1. 63. 1612

  2. 64. 2512

  3. 65. 12513

  4. 66. 2713

  5. 67. 6423

  6. 68. 2743

In Exercises 6971, simplify using properties of exponents.

  1. 69. (5x23)(4x14)

  2. 70. 15x345x12

  3. 71. (125x6)23

  4. 72. Simplify by reducing the index of the radical: y3.6

P.4

In Exercises 7374, perform the indicated operations. Write the resulting polynomial in standard form and indicate its degree.

  1. 73. (6x3+7x29x+3)+(14x3+3x211x7)

  2. 74. (13x48x3+2x2)(5x43x3+2x26)

In Exercises 7581, find each product.

  1. 75. (3x2)(4x2+3x5)

  2. 76. (3x5)(2x+1)

  3. 77. (4x+5)(4x5)

  4. 78. (2x+5)2

  5. 79. (3x4)2

  6. 80. (2x+1)3

  7. 81. (5x2)3

In Exercises 8283, perform the indicated operations. Indicate the degree of the resulting polynomial.

  1. 82. (7x28xy+y2)+(8x29xy4y2)

  2. 83. (13x3 y25x2 y9x2)(11x3 y26x2 y+3x24)

In Exercises 8488, find each product.

  1. 84. (x+7y)(3x5y)

  2. 85. (3x5y)2

  3. 86. (3x2+2y)2

  4. 87. (7x+4y)(7x4y)

  5. 88. (ab)(a2+ab+b2)

P.5

In Exercises 89105, factor completely, or state that the polynomial is prime.

  1. 89. 15x3+3x2

  2. 90. x211x+28

  3. 91. 15x2x2

  4. 92. 64x2

  5. 93. x2+16

  6. 94. 3x49x330x2

  7. 95. 20x736x3

  8. 96. x33x29x+27

  9. 97. 16x240x+25

  10. 98. x416

  11. 99. y38

  12. 100. x3+64

  13. 101. 3x412x2

  14. 102. 27x3125

  15. 103. x5x

  16. 104. x3+5x22x10

  17. 105. x2+18x+81y2

In Exercises 106108, factor and simplify each algebraic expression.

  1. 106. 16x34+32x14

  2. 107. (x24)(x2+3)12(x24)2(x2+3)32

  3. 108. 12x12+6x32

P.6

In Exercises 109111, simplify each rational expression. Also, list all numbers that must be excluded from the domain.

  1. 109. x3+2x2x+2

  2. 110. x2+3x18x236

  3. 111. x2+2xx2+4x+4

In Exercises 112114, multiply or divide as indicated.

  1. 112. x2+6x+9x24x+3x2

  2. 113. 6x+2x21÷3x2+xx1

  3. 114. x25x24x2x12÷x210x+16x2+x6

In Exercises 115120, add or subtract as indicated.

  1. 115. 2x7x29x10x29

  2. 116. x+283x

  3. 117. 1225x2+1115x3

  4. 118. 3xx+2+xx2

  5. 119. xx29+x1x25x+6

  6. 120. 4x12x2+5x3x+36x2+x2

In Exercises 121124, simplify each complex rational expression.

  1. 121. 1x1213x6

  2. 122. 3+12x116x2

  3. 123. 31x+33+1x+3

  4. 124. 25x2+x225x225x2

P.7

In Exercises 125138, solve each equation.

  1. 125. 12(6x)=3x+2

  2. 126. 2(x4)+3(x+5)=2x2

  3. 127. 2x4(5x+1)=3x+17

  4. 128. 1x11x+1=2x21

  5. 129. 4x+2+2x4=30x22x8

  6. 130. 4|2x+1|+12=0

  7. 131. 2x211x+5=0

  8. 132. (3x+5)(x3)=5

  9. 133. 3x27x+1=0

  10. 134. x29=0

  11. 135. (x3)224=0

  12. 136. 2xx2+6x+8=xx+42x+2

  13. 137. 82xx=0

  14. 138. 2x3+x=3

In Exercises 139140, solve each formula for the specified variable.

  1. 139. υt+gt2=s for g

  2. 140. T=APPr for P

In Exercises 141142, without solving the given quadratic equation, determine the number and type of solutions.

  1. 141. x2=2x19

  2. 142. 9x230x+25=0

P.8

In Exercises 143155, use the five-step strategy for solving word problems.

  1. 143. The Dog Ate My Calendar. The bar graph shows seven common excuses by college students for not meeting assignment deadlines. The bar heights represent the number of excuses for every 500 excuses that fall into each of these categories.

    A bar chart shows excuses by college students for not meeting assignment deadlines.

    Source: Roig and Caso, “Lying and Cheating: Fraudulent Excuse Making, Cheating, and Plagiarism,” Journal of Psychology

    For every 500 excuses, the number involving computer problems exceeds the number involving oversleeping by 10. The number involving illness exceeds the number involving oversleeping by 80. Combined, oversleeping, computer problems, and illness account for 270 excuses for not meeting assignment deadlines. For every 500 excuses, determine the number due to oversleeping, computer problems, and illness.

  2. 144. The bar graph shows the average price of a movie ticket for selected years from 1980 through 2019. The graph indicates that in 1980, the average movie ticket price was $2.69. For the period from 1980 through 2019, the price increased by approximately $0.17 per year. If this trend continues, by which year will the average price of a movie ticket be $10?

    A bar chart shows the average price of a U.S. movie ticket.

    Sources: Motion Picture Association of America, National Association of Theater Owners (NATO), and Bureau of Labor Statistics (BLS)

  3. 145. You are choosing between two internet service providers. The first has a one-time installation and activation fee of $150 and a monthly charge of $60. The other offers the same services with a one-time fee of $30 and a monthly charge of $75. After how many months will the total costs for the two providers be the same?

  4. 146. An apartment complex has offered you a move-in special of 30% off the first month’s rent. If you pay $945 for the first month, what should you expect to pay for the second month when you must pay full price?

  5. 147. A real estate agent receives 3% commission on the sales price of a home. The agent has incurred $2125 in advertising and other expenses listing the home. If the agent would like to earn $9125 after expenses, what sales price is necessary?

  6. 148. You invested $9000 in two funds paying 1.7% and 1.9% annual interest. At the end of the year, the total interest from these investments was $166. How much was invested at each rate?

  7. 149. Last month you had a total of $5000 in interest-bearing balances on two credit cards. One card has a monthly interest rate of 1.75%, and the other has a monthly rate of 2.25%. If your total interest for the month was $94.75, what was the interest-bearing balance on each card?

  8. 150. The length of a rectangular field is 6 yards less than triple the width. If the perimeter of the field is 340 yards, what are its dimensions?

  9. 151. In 2015, there were 14,100 students at college A, with a projected enrollment increase of 1500 students per year. In the same year, there were 41,700 students at college B, with a projected enrollment decline of 800 students per year. In which year will the colleges have the same enrollment? What will be the enrollment in each college at that time?

  10. 152. An architect is allowed 15 square yards of floor space to add a small bedroom to a house. Because of the room’s design in relationship to the existing structure, the width of the rectangular floor must be 7 yards less than two times the length. Find the length and width of the rectangular floor that the architect is permitted.

  11. 153. A building casts a shadow that is double the length of its height. If the distance from the end of the shadow to the top of the building is 300 meters, how high is the building? Round to the nearest meter.

  12. 154. A painting measuring 10 inches by 16 inches is surrounded by a frame of uniform width. If the combined area of the painting and frame is 280 square inches, determine the width of the frame.

  13. 155. Club members equally share the cost of $1500 to charter a fishing boat. Shortly before the boat is to leave, four people decide not to go due to rough seas. As a result, the cost per person is increased by $100. How many people originally intended to go on the fishing trip?

P.9

In Exercises 156158, express each interval in set-builder notation and graph the interval on a number line.

  1. 156. [3, 5)

  2. 157. (2, )

  3. 158. (, 0]

In Exercises 159162, use graphs to find each set.

  1. 159. (2, 1][1, 3)

  2. 160. (2, 1][1, 3)

  3. 161. [1, 3)(0, 4)

  4. 162. [1, 3)(0, 4)

In Exercises 163172, solve each inequality. Other than inequalities with no solution, use interval notation to express solution sets and graph each solution set on a number line.

  1. 163. 6x+315

  2. 164. 6x94x3

  3. 165. x3341>x2

  4. 166. 6x+5>2(x3)25

  5. 167. 3(2x1)2(x4)7+2(3+4x)

  6. 168. 7<2x+39

  7. 169. |2x+3|15

  8. 170. |2x+63|>2

  9. 171. |2x+5|76

  10. 172. 4|x+2|+57

  11. 173. A car rental agency rents a certain car for $40 per day with unlimited mileage or $24 per day plus $0.20 per mile. How far can a customer drive this car per day for the $24 option to cost no more than the unlimited mileage option?

  12. 174. To receive a B in a course, you must have an average of at least 80% but less than 90% on five exams. Your grades on the first four exams were 95%, 79%, 91%, and 86%. What range of grades on the fifth exam will result in a B for the course?

Chapter P: Review Exercises

Chapter P Review Exercises

You can use these review exercises, like the review exercises at the end of each chapter, to test your understanding of the chapter’s topics. However, you can also use these exercises as a prerequisite test to check your mastery of the fundamental algebra skills needed in this book.

P.1

In Exercises 12, evaluate each algebraic expression for the given value or values of the variable(s).

  1. 1. 3+6(x2)3 for x=4

  2. 2. x25(xy) for x=6 and y=2

  3. 3. You are riding along an expressway traveling x miles per hour. The formula

    S=0.015x2+x+10

    models the recommended safe distance, S, in feet, between your car and other cars on the expressway. What is the recommended safe distance when your speed is 60 miles per hour?

In Exercises 47, let A={a, b, c}, B={a, c, d, e}, and C={a, d, f, g}. Find the indicated set.

  1. 4. AB

  2. 5. AB

  3. 6. AC

  4. 7. CA

  5. 8. Consider the set:

    {17, 913, 0, 0.75, 2, π, 81}.

    List all numbers from the set that are a. natural numbers, b. whole numbers, c. integers, d. rational numbers, e. irrational numbers, f. real numbers.

In Exercises 911, rewrite each expression without absolute value bars.

  1. 9. |103|

  2. 10. |21|

  3. 11. |317|

  4. 12. Express the distance between the numbers 17 and 4 using absolute value. Then evaluate the absolute value.

In Exercises 1318, state the name of the property illustrated.

  1. 13. 3+17=17+3

  2. 14. (63)9=6(39)

  3. 15. 3(5+3)=15+3

  4. 16. (69)2=2(69)

  5. 17. 3(5+3)=(5+3)3

  6. 18. (37)+(47)=(47)+(37)

In Exercises 1922, simplify each algebraic expression.

  1. 19. 5(2x3)+7x

  2. 20. 15(5x)+[(3y)+(3y)](x)

  3. 21. 3(4y5)(7y+2)

  4. 22. 82[3(5x1)]

  5. 23. The diversity index, from 0 (no diversity) to 100, measures the chance that two randomly selected people are a different race or ethnicity. The diversity index in the United States varies widely from region to region, from as high as 81 in Hawaii to as low as 11 in Vermont. The bar graph shows the national diversity index for the United States for five years in the period from 1980 through 2020.

    A bar graph shows the national diversity index for US for five years.

    Source: USA Today

    The data in the graph can be modeled by the formula

    D=0.005x2+0.55x+34,

    where D is the national diversity index in the United States x years after 1980. According to the formula, what was the U.S. diversity index in 2010? How does this compare with the index displayed by the bar graph?

P.2

Evaluate each exponential expression in Exercises 2427.

  1. 24. (3)3(2)2

  2. 25. 24+41

  3. 26. 535

  4. 27. 3336

Simplify each exponential expression in Exercises 2831.

  1. 28. (2x4 y3)3

  2. 29. (5x3 y2)(2x11 y2)

  3. 30. (2x3)4

  4. 31. 7x5 y628x15 y2

In Exercises 3233, write each number in decimal notation.

  1. 32. 3.74×104

  2. 33. 7.45×105

In Exercises 3435, write each number in scientific notation.

  1. 34. 3,590,000

  2. 35. 0.00725

In Exercises 3637, perform the indicated operation and write the answer in decimal notation.

  1. 36. (3×103)(1.3×102)

  2. 37. 6.9×1033×105

  3. 38. The average salary of a professional baseball player is $4.1 million. (Source: Major League Baseball Player Association) Express this number in scientific notation.

  4. 39. The average salary of a nurse is $73,000. (Source: U.S. Department of Labor) Express this number in scientific notation.

  5. 40. Use your scientific notation answers from Exercises 38 and 39 to answer this question.

    How many times greater is the average salary of a professional baseball player than the average salary of a nurse?

P.3

Use the product rule to simplify the expressions in Exercises 4144. In Exercises 4344, assume that variables represent nonnegative real numbers.

  1. 41. 300

  2. 42. 12x2

  3. 43. 10x2x

  4. 44. r3

Use the quotient rule to simplify the expressions in Exercises 4546.

  1. 45. 1214

  2. 46. 96x32x (Assume that x>0.)

In Exercises 4749, add or subtract terms whenever possible.

  1. 47. 75+135

  2. 48. 250+38

  3. 49. 472248

In Exercises 5053, rationalize the denominator.

  1. 50. 305

  2. 51. 23

  3. 52. 56+3

  4. 53. 1475

Evaluate each expression in Exercises 5457 or indicate that the root is not a real number.

  1. 54. 1253

  2. 55. 325

  3. 56. 1254

  4. 57. (5)44

Simplify the radical expressions in Exercises 5862.

  1. 58. 813

  2. 59. y53

  3. 60. 84104

  4. 61. 4163+523

  5. 62. 32x5416x4 (Assume that x>0.)

In Exercises 6368, evaluate each expression.

  1. 63. 1612

  2. 64. 2512

  3. 65. 12513

  4. 66. 2713

  5. 67. 6423

  6. 68. 2743

In Exercises 6971, simplify using properties of exponents.

  1. 69. (5x23)(4x14)

  2. 70. 15x345x12

  3. 71. (125x6)23

  4. 72. Simplify by reducing the index of the radical: y3.6

P.4

In Exercises 7374, perform the indicated operations. Write the resulting polynomial in standard form and indicate its degree.

  1. 73. (6x3+7x29x+3)+(14x3+3x211x7)

  2. 74. (13x48x3+2x2)(5x43x3+2x26)

In Exercises 7581, find each product.

  1. 75. (3x2)(4x2+3x5)

  2. 76. (3x5)(2x+1)

  3. 77. (4x+5)(4x5)

  4. 78. (2x+5)2

  5. 79. (3x4)2

  6. 80. (2x+1)3

  7. 81. (5x2)3

In Exercises 8283, perform the indicated operations. Indicate the degree of the resulting polynomial.

  1. 82. (7x28xy+y2)+(8x29xy4y2)

  2. 83. (13x3 y25x2 y9x2)(11x3 y26x2 y+3x24)

In Exercises 8488, find each product.

  1. 84. (x+7y)(3x5y)

  2. 85. (3x5y)2

  3. 86. (3x2+2y)2

  4. 87. (7x+4y)(7x4y)

  5. 88. (ab)(a2+ab+b2)

P.5

In Exercises 89105, factor completely, or state that the polynomial is prime.

  1. 89. 15x3+3x2

  2. 90. x211x+28

  3. 91. 15x2x2

  4. 92. 64x2

  5. 93. x2+16

  6. 94. 3x49x330x2

  7. 95. 20x736x3

  8. 96. x33x29x+27

  9. 97. 16x240x+25

  10. 98. x416

  11. 99. y38

  12. 100. x3+64

  13. 101. 3x412x2

  14. 102. 27x3125

  15. 103. x5x

  16. 104. x3+5x22x10

  17. 105. x2+18x+81y2

In Exercises 106108, factor and simplify each algebraic expression.

  1. 106. 16x34+32x14

  2. 107. (x24)(x2+3)12(x24)2(x2+3)32

  3. 108. 12x12+6x32

P.6

In Exercises 109111, simplify each rational expression. Also, list all numbers that must be excluded from the domain.

  1. 109. x3+2x2x+2

  2. 110. x2+3x18x236

  3. 111. x2+2xx2+4x+4

In Exercises 112114, multiply or divide as indicated.

  1. 112. x2+6x+9x24x+3x2

  2. 113. 6x+2x21÷3x2+xx1

  3. 114. x25x24x2x12÷x210x+16x2+x6

In Exercises 115120, add or subtract as indicated.

  1. 115. 2x7x29x10x29

  2. 116. x+283x

  3. 117. 1225x2+1115x3

  4. 118. 3xx+2+xx2

  5. 119. xx29+x1x25x+6

  6. 120. 4x12x2+5x3x+36x2+x2

In Exercises 121124, simplify each complex rational expression.

  1. 121. 1x1213x6

  2. 122. 3+12x116x2

  3. 123. 31x+33+1x+3

  4. 124. 25x2+x225x225x2

P.7

In Exercises 125138, solve each equation.

  1. 125. 12(6x)=3x+2

  2. 126. 2(x4)+3(x+5)=2x2

  3. 127. 2x4(5x+1)=3x+17

  4. 128. 1x11x+1=2x21

  5. 129. 4x+2+2x4=30x22x8

  6. 130. 4|2x+1|+12=0

  7. 131. 2x211x+5=0

  8. 132. (3x+5)(x3)=5

  9. 133. 3x27x+1=0

  10. 134. x29=0

  11. 135. (x3)224=0

  12. 136. 2xx2+6x+8=xx+42x+2

  13. 137. 82xx=0

  14. 138. 2x3+x=3

In Exercises 139140, solve each formula for the specified variable.

  1. 139. υt+gt2=s for g

  2. 140. T=APPr for P

In Exercises 141142, without solving the given quadratic equation, determine the number and type of solutions.

  1. 141. x2=2x19

  2. 142. 9x230x+25=0

P.8

In Exercises 143155, use the five-step strategy for solving word problems.

  1. 143. The Dog Ate My Calendar. The bar graph shows seven common excuses by college students for not meeting assignment deadlines. The bar heights represent the number of excuses for every 500 excuses that fall into each of these categories.

    A bar chart shows excuses by college students for not meeting assignment deadlines.

    Source: Roig and Caso, “Lying and Cheating: Fraudulent Excuse Making, Cheating, and Plagiarism,” Journal of Psychology

    For every 500 excuses, the number involving computer problems exceeds the number involving oversleeping by 10. The number involving illness exceeds the number involving oversleeping by 80. Combined, oversleeping, computer problems, and illness account for 270 excuses for not meeting assignment deadlines. For every 500 excuses, determine the number due to oversleeping, computer problems, and illness.

  2. 144. The bar graph shows the average price of a movie ticket for selected years from 1980 through 2019. The graph indicates that in 1980, the average movie ticket price was $2.69. For the period from 1980 through 2019, the price increased by approximately $0.17 per year. If this trend continues, by which year will the average price of a movie ticket be $10?

    A bar chart shows the average price of a U.S. movie ticket.

    Sources: Motion Picture Association of America, National Association of Theater Owners (NATO), and Bureau of Labor Statistics (BLS)

  3. 145. You are choosing between two internet service providers. The first has a one-time installation and activation fee of $150 and a monthly charge of $60. The other offers the same services with a one-time fee of $30 and a monthly charge of $75. After how many months will the total costs for the two providers be the same?

  4. 146. An apartment complex has offered you a move-in special of 30% off the first month’s rent. If you pay $945 for the first month, what should you expect to pay for the second month when you must pay full price?

  5. 147. A real estate agent receives 3% commission on the sales price of a home. The agent has incurred $2125 in advertising and other expenses listing the home. If the agent would like to earn $9125 after expenses, what sales price is necessary?

  6. 148. You invested $9000 in two funds paying 1.7% and 1.9% annual interest. At the end of the year, the total interest from these investments was $166. How much was invested at each rate?

  7. 149. Last month you had a total of $5000 in interest-bearing balances on two credit cards. One card has a monthly interest rate of 1.75%, and the other has a monthly rate of 2.25%. If your total interest for the month was $94.75, what was the interest-bearing balance on each card?

  8. 150. The length of a rectangular field is 6 yards less than triple the width. If the perimeter of the field is 340 yards, what are its dimensions?

  9. 151. In 2015, there were 14,100 students at college A, with a projected enrollment increase of 1500 students per year. In the same year, there were 41,700 students at college B, with a projected enrollment decline of 800 students per year. In which year will the colleges have the same enrollment? What will be the enrollment in each college at that time?

  10. 152. An architect is allowed 15 square yards of floor space to add a small bedroom to a house. Because of the room’s design in relationship to the existing structure, the width of the rectangular floor must be 7 yards less than two times the length. Find the length and width of the rectangular floor that the architect is permitted.

  11. 153. A building casts a shadow that is double the length of its height. If the distance from the end of the shadow to the top of the building is 300 meters, how high is the building? Round to the nearest meter.

  12. 154. A painting measuring 10 inches by 16 inches is surrounded by a frame of uniform width. If the combined area of the painting and frame is 280 square inches, determine the width of the frame.

  13. 155. Club members equally share the cost of $1500 to charter a fishing boat. Shortly before the boat is to leave, four people decide not to go due to rough seas. As a result, the cost per person is increased by $100. How many people originally intended to go on the fishing trip?

P.9

In Exercises 156158, express each interval in set-builder notation and graph the interval on a number line.

  1. 156. [3, 5)

  2. 157. (2, )

  3. 158. (, 0]

In Exercises 159162, use graphs to find each set.

  1. 159. (2, 1][1, 3)

  2. 160. (2, 1][1, 3)

  3. 161. [1, 3)(0, 4)

  4. 162. [1, 3)(0, 4)

In Exercises 163172, solve each inequality. Other than inequalities with no solution, use interval notation to express solution sets and graph each solution set on a number line.

  1. 163. 6x+315

  2. 164. 6x94x3

  3. 165. x3341>x2

  4. 166. 6x+5>2(x3)25

  5. 167. 3(2x1)2(x4)7+2(3+4x)

  6. 168. 7<2x+39

  7. 169. |2x+3|15

  8. 170. |2x+63|>2

  9. 171. |2x+5|76

  10. 172. 4|x+2|+57

  11. 173. A car rental agency rents a certain car for $40 per day with unlimited mileage or $24 per day plus $0.20 per mile. How far can a customer drive this car per day for the $24 option to cost no more than the unlimited mileage option?

  12. 174. To receive a B in a course, you must have an average of at least 80% but less than 90% on five exams. Your grades on the first four exams were 95%, 79%, 91%, and 86%. What range of grades on the fifth exam will result in a B for the course?

Chapter P: Review Exercises

Chapter P Review Exercises

You can use these review exercises, like the review exercises at the end of each chapter, to test your understanding of the chapter’s topics. However, you can also use these exercises as a prerequisite test to check your mastery of the fundamental algebra skills needed in this book.

P.1

In Exercises 12, evaluate each algebraic expression for the given value or values of the variable(s).

  1. 1. 3+6(x2)3 for x=4

  2. 2. x25(xy) for x=6 and y=2

  3. 3. You are riding along an expressway traveling x miles per hour. The formula

    S=0.015x2+x+10

    models the recommended safe distance, S, in feet, between your car and other cars on the expressway. What is the recommended safe distance when your speed is 60 miles per hour?

In Exercises 47, let A={a, b, c}, B={a, c, d, e}, and C={a, d, f, g}. Find the indicated set.

  1. 4. AB

  2. 5. AB

  3. 6. AC

  4. 7. CA

  5. 8. Consider the set:

    {17, 913, 0, 0.75, 2, π, 81}.

    List all numbers from the set that are a. natural numbers, b. whole numbers, c. integers, d. rational numbers, e. irrational numbers, f. real numbers.

In Exercises 911, rewrite each expression without absolute value bars.

  1. 9. |103|

  2. 10. |21|

  3. 11. |317|

  4. 12. Express the distance between the numbers 17 and 4 using absolute value. Then evaluate the absolute value.

In Exercises 1318, state the name of the property illustrated.

  1. 13. 3+17=17+3

  2. 14. (63)9=6(39)

  3. 15. 3(5+3)=15+3

  4. 16. (69)2=2(69)

  5. 17. 3(5+3)=(5+3)3

  6. 18. (37)+(47)=(47)+(37)

In Exercises 1922, simplify each algebraic expression.

  1. 19. 5(2x3)+7x

  2. 20. 15(5x)+[(3y)+(3y)](x)

  3. 21. 3(4y5)(7y+2)

  4. 22. 82[3(5x1)]

  5. 23. The diversity index, from 0 (no diversity) to 100, measures the chance that two randomly selected people are a different race or ethnicity. The diversity index in the United States varies widely from region to region, from as high as 81 in Hawaii to as low as 11 in Vermont. The bar graph shows the national diversity index for the United States for five years in the period from 1980 through 2020.

    A bar graph shows the national diversity index for US for five years.

    Source: USA Today

    The data in the graph can be modeled by the formula

    D=0.005x2+0.55x+34,

    where D is the national diversity index in the United States x years after 1980. According to the formula, what was the U.S. diversity index in 2010? How does this compare with the index displayed by the bar graph?

P.2

Evaluate each exponential expression in Exercises 2427.

  1. 24. (3)3(2)2

  2. 25. 24+41

  3. 26. 535

  4. 27. 3336

Simplify each exponential expression in Exercises 2831.

  1. 28. (2x4 y3)3

  2. 29. (5x3 y2)(2x11 y2)

  3. 30. (2x3)4

  4. 31. 7x5 y628x15 y2

In Exercises 3233, write each number in decimal notation.

  1. 32. 3.74×104

  2. 33. 7.45×105

In Exercises 3435, write each number in scientific notation.

  1. 34. 3,590,000

  2. 35. 0.00725

In Exercises 3637, perform the indicated operation and write the answer in decimal notation.

  1. 36. (3×103)(1.3×102)

  2. 37. 6.9×1033×105

  3. 38. The average salary of a professional baseball player is $4.1 million. (Source: Major League Baseball Player Association) Express this number in scientific notation.

  4. 39. The average salary of a nurse is $73,000. (Source: U.S. Department of Labor) Express this number in scientific notation.

  5. 40. Use your scientific notation answers from Exercises 38 and 39 to answer this question.

    How many times greater is the average salary of a professional baseball player than the average salary of a nurse?

P.3

Use the product rule to simplify the expressions in Exercises 4144. In Exercises 4344, assume that variables represent nonnegative real numbers.

  1. 41. 300

  2. 42. 12x2

  3. 43. 10x2x

  4. 44. r3

Use the quotient rule to simplify the expressions in Exercises 4546.

  1. 45. 1214

  2. 46. 96x32x (Assume that x>0.)

In Exercises 4749, add or subtract terms whenever possible.

  1. 47. 75+135

  2. 48. 250+38

  3. 49. 472248

In Exercises 5053, rationalize the denominator.

  1. 50. 305

  2. 51. 23

  3. 52. 56+3

  4. 53. 1475

Evaluate each expression in Exercises 5457 or indicate that the root is not a real number.

  1. 54. 1253

  2. 55. 325

  3. 56. 1254

  4. 57. (5)44

Simplify the radical expressions in Exercises 5862.

  1. 58. 813

  2. 59. y53

  3. 60. 84104

  4. 61. 4163+523

  5. 62. 32x5416x4 (Assume that x>0.)

In Exercises 6368, evaluate each expression.

  1. 63. 1612

  2. 64. 2512

  3. 65. 12513

  4. 66. 2713

  5. 67. 6423

  6. 68. 2743

In Exercises 6971, simplify using properties of exponents.

  1. 69. (5x23)(4x14)

  2. 70. 15x345x12

  3. 71. (125x6)23

  4. 72. Simplify by reducing the index of the radical: y3.6

P.4

In Exercises 7374, perform the indicated operations. Write the resulting polynomial in standard form and indicate its degree.

  1. 73. (6x3+7x29x+3)+(14x3+3x211x7)

  2. 74. (13x48x3+2x2)(5x43x3+2x26)

In Exercises 7581, find each product.

  1. 75. (3x2)(4x2+3x5)

  2. 76. (3x5)(2x+1)

  3. 77. (4x+5)(4x5)

  4. 78. (2x+5)2

  5. 79. (3x4)2

  6. 80. (2x+1)3

  7. 81. (5x2)3

In Exercises 8283, perform the indicated operations. Indicate the degree of the resulting polynomial.

  1. 82. (7x28xy+y2)+(8x29xy4y2)

  2. 83. (13x3 y25x2 y9x2)(11x3 y26x2 y+3x24)

In Exercises 8488, find each product.

  1. 84. (x+7y)(3x5y)

  2. 85. (3x5y)2

  3. 86. (3x2+2y)2

  4. 87. (7x+4y)(7x4y)

  5. 88. (ab)(a2+ab+b2)

P.5

In Exercises 89105, factor completely, or state that the polynomial is prime.

  1. 89. 15x3+3x2

  2. 90. x211x+28

  3. 91. 15x2x2

  4. 92. 64x2

  5. 93. x2+16

  6. 94. 3x49x330x2

  7. 95. 20x736x3

  8. 96. x33x29x+27

  9. 97. 16x240x+25

  10. 98. x416

  11. 99. y38

  12. 100. x3+64

  13. 101. 3x412x2

  14. 102. 27x3125

  15. 103. x5x

  16. 104. x3+5x22x10

  17. 105. x2+18x+81y2

In Exercises 106108, factor and simplify each algebraic expression.

  1. 106. 16x34+32x14

  2. 107. (x24)(x2+3)12(x24)2(x2+3)32

  3. 108. 12x12+6x32

P.6

In Exercises 109111, simplify each rational expression. Also, list all numbers that must be excluded from the domain.

  1. 109. x3+2x2x+2

  2. 110. x2+3x18x236

  3. 111. x2+2xx2+4x+4

In Exercises 112114, multiply or divide as indicated.

  1. 112. x2+6x+9x24x+3x2

  2. 113. 6x+2x21÷3x2+xx1

  3. 114. x25x24x2x12÷x210x+16x2+x6

In Exercises 115120, add or subtract as indicated.

  1. 115. 2x7x29x10x29

  2. 116. x+283x

  3. 117. 1225x2+1115x3

  4. 118. 3xx+2+xx2

  5. 119. xx29+x1x25x+6

  6. 120. 4x12x2+5x3x+36x2+x2

In Exercises 121124, simplify each complex rational expression.

  1. 121. 1x1213x6

  2. 122. 3+12x116x2

  3. 123. 31x+33+1x+3

  4. 124. 25x2+x225x225x2

P.7

In Exercises 125138, solve each equation.

  1. 125. 12(6x)=3x+2

  2. 126. 2(x4)+3(x+5)=2x2

  3. 127. 2x4(5x+1)=3x+17

  4. 128. 1x11x+1=2x21

  5. 129. 4x+2+2x4=30x22x8

  6. 130. 4|2x+1|+12=0

  7. 131. 2x211x+5=0

  8. 132. (3x+5)(x3)=5

  9. 133. 3x27x+1=0

  10. 134. x29=0

  11. 135. (x3)224=0

  12. 136. 2xx2+6x+8=xx+42x+2

  13. 137. 82xx=0

  14. 138. 2x3+x=3

In Exercises 139140, solve each formula for the specified variable.

  1. 139. υt+gt2=s for g

  2. 140. T=APPr for P

In Exercises 141142, without solving the given quadratic equation, determine the number and type of solutions.

  1. 141. x2=2x19

  2. 142. 9x230x+25=0

P.8

In Exercises 143155, use the five-step strategy for solving word problems.

  1. 143. The Dog Ate My Calendar. The bar graph shows seven common excuses by college students for not meeting assignment deadlines. The bar heights represent the number of excuses for every 500 excuses that fall into each of these categories.

    A bar chart shows excuses by college students for not meeting assignment deadlines.

    Source: Roig and Caso, “Lying and Cheating: Fraudulent Excuse Making, Cheating, and Plagiarism,” Journal of Psychology

    For every 500 excuses, the number involving computer problems exceeds the number involving oversleeping by 10. The number involving illness exceeds the number involving oversleeping by 80. Combined, oversleeping, computer problems, and illness account for 270 excuses for not meeting assignment deadlines. For every 500 excuses, determine the number due to oversleeping, computer problems, and illness.

  2. 144. The bar graph shows the average price of a movie ticket for selected years from 1980 through 2019. The graph indicates that in 1980, the average movie ticket price was $2.69. For the period from 1980 through 2019, the price increased by approximately $0.17 per year. If this trend continues, by which year will the average price of a movie ticket be $10?

    A bar chart shows the average price of a U.S. movie ticket.

    Sources: Motion Picture Association of America, National Association of Theater Owners (NATO), and Bureau of Labor Statistics (BLS)

  3. 145. You are choosing between two internet service providers. The first has a one-time installation and activation fee of $150 and a monthly charge of $60. The other offers the same services with a one-time fee of $30 and a monthly charge of $75. After how many months will the total costs for the two providers be the same?

  4. 146. An apartment complex has offered you a move-in special of 30% off the first month’s rent. If you pay $945 for the first month, what should you expect to pay for the second month when you must pay full price?

  5. 147. A real estate agent receives 3% commission on the sales price of a home. The agent has incurred $2125 in advertising and other expenses listing the home. If the agent would like to earn $9125 after expenses, what sales price is necessary?

  6. 148. You invested $9000 in two funds paying 1.7% and 1.9% annual interest. At the end of the year, the total interest from these investments was $166. How much was invested at each rate?

  7. 149. Last month you had a total of $5000 in interest-bearing balances on two credit cards. One card has a monthly interest rate of 1.75%, and the other has a monthly rate of 2.25%. If your total interest for the month was $94.75, what was the interest-bearing balance on each card?

  8. 150. The length of a rectangular field is 6 yards less than triple the width. If the perimeter of the field is 340 yards, what are its dimensions?

  9. 151. In 2015, there were 14,100 students at college A, with a projected enrollment increase of 1500 students per year. In the same year, there were 41,700 students at college B, with a projected enrollment decline of 800 students per year. In which year will the colleges have the same enrollment? What will be the enrollment in each college at that time?

  10. 152. An architect is allowed 15 square yards of floor space to add a small bedroom to a house. Because of the room’s design in relationship to the existing structure, the width of the rectangular floor must be 7 yards less than two times the length. Find the length and width of the rectangular floor that the architect is permitted.

  11. 153. A building casts a shadow that is double the length of its height. If the distance from the end of the shadow to the top of the building is 300 meters, how high is the building? Round to the nearest meter.

  12. 154. A painting measuring 10 inches by 16 inches is surrounded by a frame of uniform width. If the combined area of the painting and frame is 280 square inches, determine the width of the frame.

  13. 155. Club members equally share the cost of $1500 to charter a fishing boat. Shortly before the boat is to leave, four people decide not to go due to rough seas. As a result, the cost per person is increased by $100. How many people originally intended to go on the fishing trip?

P.9

In Exercises 156158, express each interval in set-builder notation and graph the interval on a number line.

  1. 156. [3, 5)

  2. 157. (2, )

  3. 158. (, 0]

In Exercises 159162, use graphs to find each set.

  1. 159. (2, 1][1, 3)

  2. 160. (2, 1][1, 3)

  3. 161. [1, 3)(0, 4)

  4. 162. [1, 3)(0, 4)

In Exercises 163172, solve each inequality. Other than inequalities with no solution, use interval notation to express solution sets and graph each solution set on a number line.

  1. 163. 6x+315

  2. 164. 6x94x3

  3. 165. x3341>x2

  4. 166. 6x+5>2(x3)25

  5. 167. 3(2x1)2(x4)7+2(3+4x)

  6. 168. 7<2x+39

  7. 169. |2x+3|15

  8. 170. |2x+63|>2

  9. 171. |2x+5|76

  10. 172. 4|x+2|+57

  11. 173. A car rental agency rents a certain car for $40 per day with unlimited mileage or $24 per day plus $0.20 per mile. How far can a customer drive this car per day for the $24 option to cost no more than the unlimited mileage option?

  12. 174. To receive a B in a course, you must have an average of at least 80% but less than 90% on five exams. Your grades on the first four exams were 95%, 79%, 91%, and 86%. What range of grades on the fifth exam will result in a B for the course?

Chapter P: Review Exercises

Chapter P Review Exercises

You can use these review exercises, like the review exercises at the end of each chapter, to test your understanding of the chapter’s topics. However, you can also use these exercises as a prerequisite test to check your mastery of the fundamental algebra skills needed in this book.

P.1

In Exercises 12, evaluate each algebraic expression for the given value or values of the variable(s).

  1. 1. 3+6(x2)3 for x=4

  2. 2. x25(xy) for x=6 and y=2

  3. 3. You are riding along an expressway traveling x miles per hour. The formula

    S=0.015x2+x+10

    models the recommended safe distance, S, in feet, between your car and other cars on the expressway. What is the recommended safe distance when your speed is 60 miles per hour?

In Exercises 47, let A={a, b, c}, B={a, c, d, e}, and C={a, d, f, g}. Find the indicated set.

  1. 4. AB

  2. 5. AB

  3. 6. AC

  4. 7. CA

  5. 8. Consider the set:

    {17, 913, 0, 0.75, 2, π, 81}.

    List all numbers from the set that are a. natural numbers, b. whole numbers, c. integers, d. rational numbers, e. irrational numbers, f. real numbers.

In Exercises 911, rewrite each expression without absolute value bars.

  1. 9. |103|

  2. 10. |21|

  3. 11. |317|

  4. 12. Express the distance between the numbers 17 and 4 using absolute value. Then evaluate the absolute value.

In Exercises 1318, state the name of the property illustrated.

  1. 13. 3+17=17+3

  2. 14. (63)9=6(39)

  3. 15. 3(5+3)=15+3

  4. 16. (69)2=2(69)

  5. 17. 3(5+3)=(5+3)3

  6. 18. (37)+(47)=(47)+(37)

In Exercises 1922, simplify each algebraic expression.

  1. 19. 5(2x3)+7x

  2. 20. 15(5x)+[(3y)+(3y)](x)

  3. 21. 3(4y5)(7y+2)

  4. 22. 82[3(5x1)]

  5. 23. The diversity index, from 0 (no diversity) to 100, measures the chance that two randomly selected people are a different race or ethnicity. The diversity index in the United States varies widely from region to region, from as high as 81 in Hawaii to as low as 11 in Vermont. The bar graph shows the national diversity index for the United States for five years in the period from 1980 through 2020.

    A bar graph shows the national diversity index for US for five years.

    Source: USA Today

    The data in the graph can be modeled by the formula

    D=0.005x2+0.55x+34,

    where D is the national diversity index in the United States x years after 1980. According to the formula, what was the U.S. diversity index in 2010? How does this compare with the index displayed by the bar graph?

P.2

Evaluate each exponential expression in Exercises 2427.

  1. 24. (3)3(2)2

  2. 25. 24+41

  3. 26. 535

  4. 27. 3336

Simplify each exponential expression in Exercises 2831.

  1. 28. (2x4 y3)3

  2. 29. (5x3 y2)(2x11 y2)

  3. 30. (2x3)4

  4. 31. 7x5 y628x15 y2

In Exercises 3233, write each number in decimal notation.

  1. 32. 3.74×104

  2. 33. 7.45×105

In Exercises 3435, write each number in scientific notation.

  1. 34. 3,590,000

  2. 35. 0.00725

In Exercises 3637, perform the indicated operation and write the answer in decimal notation.

  1. 36. (3×103)(1.3×102)

  2. 37. 6.9×1033×105

  3. 38. The average salary of a professional baseball player is $4.1 million. (Source: Major League Baseball Player Association) Express this number in scientific notation.

  4. 39. The average salary of a nurse is $73,000. (Source: U.S. Department of Labor) Express this number in scientific notation.

  5. 40. Use your scientific notation answers from Exercises 38 and 39 to answer this question.

    How many times greater is the average salary of a professional baseball player than the average salary of a nurse?

P.3

Use the product rule to simplify the expressions in Exercises 4144. In Exercises 4344, assume that variables represent nonnegative real numbers.

  1. 41. 300

  2. 42. 12x2

  3. 43. 10x2x

  4. 44. r3

Use the quotient rule to simplify the expressions in Exercises 4546.

  1. 45. 1214

  2. 46. 96x32x (Assume that x>0.)

In Exercises 4749, add or subtract terms whenever possible.

  1. 47. 75+135

  2. 48. 250+38

  3. 49. 472248

In Exercises 5053, rationalize the denominator.

  1. 50. 305

  2. 51. 23

  3. 52. 56+3

  4. 53. 1475

Evaluate each expression in Exercises 5457 or indicate that the root is not a real number.

  1. 54. 1253

  2. 55. 325

  3. 56. 1254

  4. 57. (5)44

Simplify the radical expressions in Exercises 5862.

  1. 58. 813

  2. 59. y53

  3. 60. 84104

  4. 61. 4163+523

  5. 62. 32x5416x4 (Assume that x>0.)

In Exercises 6368, evaluate each expression.

  1. 63. 1612

  2. 64. 2512

  3. 65. 12513

  4. 66. 2713

  5. 67. 6423

  6. 68. 2743

In Exercises 6971, simplify using properties of exponents.

  1. 69. (5x23)(4x14)

  2. 70. 15x345x12

  3. 71. (125x6)23

  4. 72. Simplify by reducing the index of the radical: y3.6

P.4

In Exercises 7374, perform the indicated operations. Write the resulting polynomial in standard form and indicate its degree.

  1. 73. (6x3+7x29x+3)+(14x3+3x211x7)

  2. 74. (13x48x3+2x2)(5x43x3+2x26)

In Exercises 7581, find each product.

  1. 75. (3x2)(4x2+3x5)

  2. 76. (3x5)(2x+1)

  3. 77. (4x+5)(4x5)

  4. 78. (2x+5)2

  5. 79. (3x4)2

  6. 80. (2x+1)3

  7. 81. (5x2)3

In Exercises 8283, perform the indicated operations. Indicate the degree of the resulting polynomial.

  1. 82. (7x28xy+y2)+(8x29xy4y2)

  2. 83. (13x3 y25x2 y9x2)(11x3 y26x2 y+3x24)

In Exercises 8488, find each product.

  1. 84. (x+7y)(3x5y)

  2. 85. (3x5y)2

  3. 86. (3x2+2y)2

  4. 87. (7x+4y)(7x4y)

  5. 88. (ab)(a2+ab+b2)

P.5

In Exercises 89105, factor completely, or state that the polynomial is prime.

  1. 89. 15x3+3x2

  2. 90. x211x+28

  3. 91. 15x2x2

  4. 92. 64x2

  5. 93. x2+16

  6. 94. 3x49x330x2

  7. 95. 20x736x3

  8. 96. x33x29x+27

  9. 97. 16x240x+25

  10. 98. x416

  11. 99. y38

  12. 100. x3+64

  13. 101. 3x412x2

  14. 102. 27x3125

  15. 103. x5x

  16. 104. x3+5x22x10

  17. 105. x2+18x+81y2

In Exercises 106108, factor and simplify each algebraic expression.

  1. 106. 16x34+32x14

  2. 107. (x24)(x2+3)12(x24)2(x2+3)32

  3. 108. 12x12+6x32

P.6

In Exercises 109111, simplify each rational expression. Also, list all numbers that must be excluded from the domain.

  1. 109. x3+2x2x+2

  2. 110. x2+3x18x236

  3. 111. x2+2xx2+4x+4

In Exercises 112114, multiply or divide as indicated.

  1. 112. x2+6x+9x24x+3x2

  2. 113. 6x+2x21÷3x2+xx1

  3. 114. x25x24x2x12÷x210x+16x2+x6

In Exercises 115120, add or subtract as indicated.

  1. 115. 2x7x29x10x29

  2. 116. x+283x

  3. 117. 1225x2+1115x3

  4. 118. 3xx+2+xx2

  5. 119. xx29+x1x25x+6

  6. 120. 4x12x2+5x3x+36x2+x2

In Exercises 121124, simplify each complex rational expression.

  1. 121. 1x1213x6

  2. 122. 3+12x116x2

  3. 123. 31x+33+1x+3

  4. 124. 25x2+x225x225x2

P.7

In Exercises 125138, solve each equation.

  1. 125. 12(6x)=3x+2

  2. 126. 2(x4)+3(x+5)=2x2

  3. 127. 2x4(5x+1)=3x+17

  4. 128. 1x11x+1=2x21

  5. 129. 4x+2+2x4=30x22x8

  6. 130. 4|2x+1|+12=0

  7. 131. 2x211x+5=0

  8. 132. (3x+5)(x3)=5

  9. 133. 3x27x+1=0

  10. 134. x29=0

  11. 135. (x3)224=0

  12. 136. 2xx2+6x+8=xx+42x+2

  13. 137. 82xx=0

  14. 138. 2x3+x=3

In Exercises 139140, solve each formula for the specified variable.

  1. 139. υt+gt2=s for g

  2. 140. T=APPr for P

In Exercises 141142, without solving the given quadratic equation, determine the number and type of solutions.

  1. 141. x2=2x19

  2. 142. 9x230x+25=0

P.8

In Exercises 143155, use the five-step strategy for solving word problems.

  1. 143. The Dog Ate My Calendar. The bar graph shows seven common excuses by college students for not meeting assignment deadlines. The bar heights represent the number of excuses for every 500 excuses that fall into each of these categories.

    A bar chart shows excuses by college students for not meeting assignment deadlines.

    Source: Roig and Caso, “Lying and Cheating: Fraudulent Excuse Making, Cheating, and Plagiarism,” Journal of Psychology

    For every 500 excuses, the number involving computer problems exceeds the number involving oversleeping by 10. The number involving illness exceeds the number involving oversleeping by 80. Combined, oversleeping, computer problems, and illness account for 270 excuses for not meeting assignment deadlines. For every 500 excuses, determine the number due to oversleeping, computer problems, and illness.

  2. 144. The bar graph shows the average price of a movie ticket for selected years from 1980 through 2019. The graph indicates that in 1980, the average movie ticket price was $2.69. For the period from 1980 through 2019, the price increased by approximately $0.17 per year. If this trend continues, by which year will the average price of a movie ticket be $10?

    A bar chart shows the average price of a U.S. movie ticket.

    Sources: Motion Picture Association of America, National Association of Theater Owners (NATO), and Bureau of Labor Statistics (BLS)

  3. 145. You are choosing between two internet service providers. The first has a one-time installation and activation fee of $150 and a monthly charge of $60. The other offers the same services with a one-time fee of $30 and a monthly charge of $75. After how many months will the total costs for the two providers be the same?

  4. 146. An apartment complex has offered you a move-in special of 30% off the first month’s rent. If you pay $945 for the first month, what should you expect to pay for the second month when you must pay full price?

  5. 147. A real estate agent receives 3% commission on the sales price of a home. The agent has incurred $2125 in advertising and other expenses listing the home. If the agent would like to earn $9125 after expenses, what sales price is necessary?

  6. 148. You invested $9000 in two funds paying 1.7% and 1.9% annual interest. At the end of the year, the total interest from these investments was $166. How much was invested at each rate?

  7. 149. Last month you had a total of $5000 in interest-bearing balances on two credit cards. One card has a monthly interest rate of 1.75%, and the other has a monthly rate of 2.25%. If your total interest for the month was $94.75, what was the interest-bearing balance on each card?

  8. 150. The length of a rectangular field is 6 yards less than triple the width. If the perimeter of the field is 340 yards, what are its dimensions?

  9. 151. In 2015, there were 14,100 students at college A, with a projected enrollment increase of 1500 students per year. In the same year, there were 41,700 students at college B, with a projected enrollment decline of 800 students per year. In which year will the colleges have the same enrollment? What will be the enrollment in each college at that time?

  10. 152. An architect is allowed 15 square yards of floor space to add a small bedroom to a house. Because of the room’s design in relationship to the existing structure, the width of the rectangular floor must be 7 yards less than two times the length. Find the length and width of the rectangular floor that the architect is permitted.

  11. 153. A building casts a shadow that is double the length of its height. If the distance from the end of the shadow to the top of the building is 300 meters, how high is the building? Round to the nearest meter.

  12. 154. A painting measuring 10 inches by 16 inches is surrounded by a frame of uniform width. If the combined area of the painting and frame is 280 square inches, determine the width of the frame.

  13. 155. Club members equally share the cost of $1500 to charter a fishing boat. Shortly before the boat is to leave, four people decide not to go due to rough seas. As a result, the cost per person is increased by $100. How many people originally intended to go on the fishing trip?

P.9

In Exercises 156158, express each interval in set-builder notation and graph the interval on a number line.

  1. 156. [3, 5)

  2. 157. (2, )

  3. 158. (, 0]

In Exercises 159162, use graphs to find each set.

  1. 159. (2, 1][1, 3)

  2. 160. (2, 1][1, 3)

  3. 161. [1, 3)(0, 4)

  4. 162. [1, 3)(0, 4)

In Exercises 163172, solve each inequality. Other than inequalities with no solution, use interval notation to express solution sets and graph each solution set on a number line.

  1. 163. 6x+315

  2. 164. 6x94x3

  3. 165. x3341>x2

  4. 166. 6x+5>2(x3)25

  5. 167. 3(2x1)2(x4)7+2(3+4x)

  6. 168. 7<2x+39

  7. 169. |2x+3|15

  8. 170. |2x+63|>2

  9. 171. |2x+5|76

  10. 172. 4|x+2|+57

  11. 173. A car rental agency rents a certain car for $40 per day with unlimited mileage or $24 per day plus $0.20 per mile. How far can a customer drive this car per day for the $24 option to cost no more than the unlimited mileage option?

  12. 174. To receive a B in a course, you must have an average of at least 80% but less than 90% on five exams. Your grades on the first four exams were 95%, 79%, 91%, and 86%. What range of grades on the fifth exam will result in a B for the course?

Chapter P: Test

Chapter P Test

You can check your answers against those at the back of the book. Step-by-step solutions are found in the Chapter Test Prep Videos available in MyLab Math and at youtube.com/user/pearsonmathstats (playlist “Blitzer Precalculus 7e”).

In Exercises 118, simplify the given expression or perform the indicated operation (and simplify, if possible), whichever is appropriate.

  1. 1. 5(2x26x)(4x23x)

  2. 2. 7+2[3(x+1)2(3x1)]

  3. 3. {1, 2, 5}{5, a}

  4. 4. {1, 2, 5}{5, a}

  5. 5. 30x3 y46x9 y4

  6. 6. 6r3r (Assume that r0.)

  7. 7. 450318

  8. 8. 35+2

  9. 9. 16x43

  10. 10. x2+2x3x23x+2

  11. 11. 5×10620×108 (Express the answer in scientific notation.)

  12. 12. (2x5)(x24x+3)

  13. 13. (5x+3y)2

  14. 14. 2x+8x3÷x2+5x+4x29

  15. 15. xx+3+5x3

  16. 16. 2x+3x27x+122x3

  17. 17. 1xx+21+1x

  18. 18. 2xx2+52x3x2+5x2+5

In Exercises 1924, factor completely, or state that the polynomial is prime.

  1. 19. x29x+18

  2. 20. x3+2x2+3x+6

  3. 21. 25x29

  4. 22. 36x284x+49

  5. 23. y3125

  6. 24. x2+10x+259y2

  7. 25. Factor and simplify:

    x(x+3)35+(x+3)25.
  8. 26. List all the rational numbers in this set:

    {7,45,0,0.25,3,4,227,π}.

In Exercises 2728, state the name of the property illustrated.

  1. 27. 3(2+5)=3(5+2)

  2. 28. 6(7+4)=67+64

  3. 29. Express in scientific notation: 0.00076.

  4. 30. Evaluate: 2753.

  5. 31. The human body contains approximately 3.2×104 microliters of blood for every pound of body weight. Each microliter of blood contains approximately 5×106 red blood cells. Express in scientific notation the approximate number of red blood cells in the body of a 180-pound person.

  6. 32. Big (Lack of) Men on Campus In 2007, 135 women received bachelor’s degrees for every 100 men. According to the U.S. Department of Education, that gender imbalance has widened, as shown by the bar graph.

    A bar graph shows the percentage of Bachelor’s degrees awarded to U S men and women.

    Source: U.S. Department of Education

    The data for bachelor’s degrees can be described by the following mathematical models:

    The image shows the data for bachelor’s degrees.
    1. According to the first formula, what percentage of bachelor’s degrees were awarded to men in 2003? Does this underestimate or overestimate the actual percent shown by the bar graph? By how much?

    2. Use the given formulas to write a new formula with a rational expression that models the ratio of the percentage of bachelor’s degrees received by men to the percentage received by women n years after 1989. Name this new mathematical model R, for ratio.

    3. Use the formula for R to find the ratio of bachelor’s degrees received by men to degrees received by women in 2017. According to the model, how many women received bachelor’s degrees for every two men in 2014? How well does this describe the data shown by the graph?

In Exercises 3347, solve each equation or inequality. Use interval notation to express solution sets of inequalities and graph these solution sets on a number line.

  1. 33. 7(x2)=4(x+1)21

  2. 34. 2x34=x42x+14

  3. 35. 2x34x+3=8x29

  4. 36. 2x23x2=0

  5. 37. (3x1)2=75

  6. 38. x(x2)=4

  7. 39. x3+5=x

  8. 40. 82xx=0

  9. 41. |23 x6|=2

  10. 42. 3|4x7|+15=0

  11. 43. 2xx2+6x+8+2x+2=xx+4

  12. 44. 3(x+4)5x12

  13. 45. x6+18x234

  14. 46. 32x+53<6

  15. 47. |3x+2|3

In Exercises 4850, solve each formula for the specified variable.

  1. 48. V=13 lwh for h

  2. 49. yy1=m(xx1) for x

  3. 50. R=asa+s for a

    The graphs show the amount being paid in Social Security benefits and the amount going into the system. All data are expressed in billions of dollars. Amounts from 2016 through 2024 are projections.

    A line graph shows social insecurity, income, and outflow of the social security system.

    Source: 2004 Social Security Trustees Report

    Exercises 5153 are based on the data shown by the graphs.

  4. 51. In 2004, the system’s income was $575 billion, projected to increase at an average rate of $43 billion per year. In which year was the system’s income $1177 billion?

  5. 52. The data for the system’s outflow can be modeled by the formula

    B=0.07x2+47.4x+500,

    where B represents the amount paid in benefits, in billions of dollars, x years after 2004. According to this model, when was the amount paid in benefits $1177 billion? Round to the nearest year.

  6. 53. How well do your answers to Exercises 51 and 52 model the data shown by the graphs?

  7. 54. Here’s Looking at You. According to University of Texas economist Daniel Hamermesh (Beauty Pays: Why Attractive People Are More Successful), strikingly attractive and good-looking men and women can expect to earn an average of $230,000 more in a lifetime than a person who is homely or plain. The bar graph shows the distribution of looks for American men and women, ranging from homely to strikingly attractive.

    A bar chart shows the distribution of looks of men and women in the United States.

    Source: Time, August 22, 2011

    The percentage of average-looking men exceeds the percentage of strikingly attractive men by 57. The percentage of good-looking men exceeds the percentage of strikingly attractive men by 25. A total of 88% of American men range between average-looking, good-looking, and strikingly attractive. Find the percentage of men who fall within each of these three categories of looks.

  8. 55. The costs for two different kinds of heating systems for a small home are given in the following table. After how many years will total costs for solar heating and electric heating be the same? What will be the cost at that time?

    System Cost to Install Operating Cost/Year
    Solar $29,700 $150
    Electric $5000 $1100
  9. 56. You invested $10,000 in two accounts paying 1.3% and 1.7% annual interest. At the end of the year, the total interest from these investments was $158. How much was invested at each rate?

  10. 57. The length of a rectangular carpet is 4 feet greater than twice its width. If the area is 48 square feet, find the carpet’s length and width.

  11. 58. A vertical pole is to be supported by a wire that is 26 feet long and anchored 24 feet from the base of the pole. How far up the pole should the wire be attached?

  12. 59. After a 60% reduction, a jacket sold for $52. What was the jacket’s price before the reduction?

  13. 60. A group of people would like to buy a vacation cabin for $600,000, sharing the cost equally. If they could find five more people to join them, each person’s share would be reduced by $6000. How many people are in the group?

  14. 61. You are choosing between two gyms. The first has a one-time membership fee of $30 and charges $10 per month. The second has a one-time fee of $10 and charges $15 per month. How many months of membership make the second gym the better deal?

Chapter P: Test

Chapter P Test

You can check your answers against those at the back of the book. Step-by-step solutions are found in the Chapter Test Prep Videos available in MyLab Math and at youtube.com/user/pearsonmathstats (playlist “Blitzer Precalculus 7e”).

In Exercises 118, simplify the given expression or perform the indicated operation (and simplify, if possible), whichever is appropriate.

  1. 1. 5(2x26x)(4x23x)

  2. 2. 7+2[3(x+1)2(3x1)]

  3. 3. {1, 2, 5}{5, a}

  4. 4. {1, 2, 5}{5, a}

  5. 5. 30x3 y46x9 y4

  6. 6. 6r3r (Assume that r0.)

  7. 7. 450318

  8. 8. 35+2

  9. 9. 16x43

  10. 10. x2+2x3x23x+2

  11. 11. 5×10620×108 (Express the answer in scientific notation.)

  12. 12. (2x5)(x24x+3)

  13. 13. (5x+3y)2

  14. 14. 2x+8x3÷x2+5x+4x29

  15. 15. xx+3+5x3

  16. 16. 2x+3x27x+122x3

  17. 17. 1xx+21+1x

  18. 18. 2xx2+52x3x2+5x2+5

In Exercises 1924, factor completely, or state that the polynomial is prime.

  1. 19. x29x+18

  2. 20. x3+2x2+3x+6

  3. 21. 25x29

  4. 22. 36x284x+49

  5. 23. y3125

  6. 24. x2+10x+259y2

  7. 25. Factor and simplify:

    x(x+3)35+(x+3)25.
  8. 26. List all the rational numbers in this set:

    {7,45,0,0.25,3,4,227,π}.

In Exercises 2728, state the name of the property illustrated.

  1. 27. 3(2+5)=3(5+2)

  2. 28. 6(7+4)=67+64

  3. 29. Express in scientific notation: 0.00076.

  4. 30. Evaluate: 2753.

  5. 31. The human body contains approximately 3.2×104 microliters of blood for every pound of body weight. Each microliter of blood contains approximately 5×106 red blood cells. Express in scientific notation the approximate number of red blood cells in the body of a 180-pound person.

  6. 32. Big (Lack of) Men on Campus In 2007, 135 women received bachelor’s degrees for every 100 men. According to the U.S. Department of Education, that gender imbalance has widened, as shown by the bar graph.

    A bar graph shows the percentage of Bachelor’s degrees awarded to U S men and women.

    Source: U.S. Department of Education

    The data for bachelor’s degrees can be described by the following mathematical models:

    The image shows the data for bachelor’s degrees.
    1. According to the first formula, what percentage of bachelor’s degrees were awarded to men in 2003? Does this underestimate or overestimate the actual percent shown by the bar graph? By how much?

    2. Use the given formulas to write a new formula with a rational expression that models the ratio of the percentage of bachelor’s degrees received by men to the percentage received by women n years after 1989. Name this new mathematical model R, for ratio.

    3. Use the formula for R to find the ratio of bachelor’s degrees received by men to degrees received by women in 2017. According to the model, how many women received bachelor’s degrees for every two men in 2014? How well does this describe the data shown by the graph?

In Exercises 3347, solve each equation or inequality. Use interval notation to express solution sets of inequalities and graph these solution sets on a number line.

  1. 33. 7(x2)=4(x+1)21

  2. 34. 2x34=x42x+14

  3. 35. 2x34x+3=8x29

  4. 36. 2x23x2=0

  5. 37. (3x1)2=75

  6. 38. x(x2)=4

  7. 39. x3+5=x

  8. 40. 82xx=0

  9. 41. |23 x6|=2

  10. 42. 3|4x7|+15=0

  11. 43. 2xx2+6x+8+2x+2=xx+4

  12. 44. 3(x+4)5x12

  13. 45. x6+18x234

  14. 46. 32x+53<6

  15. 47. |3x+2|3

In Exercises 4850, solve each formula for the specified variable.

  1. 48. V=13 lwh for h

  2. 49. yy1=m(xx1) for x

  3. 50. R=asa+s for a

    The graphs show the amount being paid in Social Security benefits and the amount going into the system. All data are expressed in billions of dollars. Amounts from 2016 through 2024 are projections.

    A line graph shows social insecurity, income, and outflow of the social security system.

    Source: 2004 Social Security Trustees Report

    Exercises 5153 are based on the data shown by the graphs.

  4. 51. In 2004, the system’s income was $575 billion, projected to increase at an average rate of $43 billion per year. In which year was the system’s income $1177 billion?

  5. 52. The data for the system’s outflow can be modeled by the formula

    B=0.07x2+47.4x+500,

    where B represents the amount paid in benefits, in billions of dollars, x years after 2004. According to this model, when was the amount paid in benefits $1177 billion? Round to the nearest year.

  6. 53. How well do your answers to Exercises 51 and 52 model the data shown by the graphs?

  7. 54. Here’s Looking at You. According to University of Texas economist Daniel Hamermesh (Beauty Pays: Why Attractive People Are More Successful), strikingly attractive and good-looking men and women can expect to earn an average of $230,000 more in a lifetime than a person who is homely or plain. The bar graph shows the distribution of looks for American men and women, ranging from homely to strikingly attractive.

    A bar chart shows the distribution of looks of men and women in the United States.

    Source: Time, August 22, 2011

    The percentage of average-looking men exceeds the percentage of strikingly attractive men by 57. The percentage of good-looking men exceeds the percentage of strikingly attractive men by 25. A total of 88% of American men range between average-looking, good-looking, and strikingly attractive. Find the percentage of men who fall within each of these three categories of looks.

  8. 55. The costs for two different kinds of heating systems for a small home are given in the following table. After how many years will total costs for solar heating and electric heating be the same? What will be the cost at that time?

    System Cost to Install Operating Cost/Year
    Solar $29,700 $150
    Electric $5000 $1100
  9. 56. You invested $10,000 in two accounts paying 1.3% and 1.7% annual interest. At the end of the year, the total interest from these investments was $158. How much was invested at each rate?

  10. 57. The length of a rectangular carpet is 4 feet greater than twice its width. If the area is 48 square feet, find the carpet’s length and width.

  11. 58. A vertical pole is to be supported by a wire that is 26 feet long and anchored 24 feet from the base of the pole. How far up the pole should the wire be attached?

  12. 59. After a 60% reduction, a jacket sold for $52. What was the jacket’s price before the reduction?

  13. 60. A group of people would like to buy a vacation cabin for $600,000, sharing the cost equally. If they could find five more people to join them, each person’s share would be reduced by $6000. How many people are in the group?

  14. 61. You are choosing between two gyms. The first has a one-time membership fee of $30 and charges $10 per month. The second has a one-time fee of $10 and charges $15 per month. How many months of membership make the second gym the better deal?

Chapter 1: Functions and Graphs

Chapter 1 Functions and Graphs

A vast expanse of open water at the top of our world was once covered with ice. The melting of the Arctic ice caps has forced polar bears to swim as far as 40 miles, causing them to drown in significant numbers. Such deaths were rare in the past.

There is strong scientific consensus that human activities are changing the Earth’s climate. Scientists now believe that there is a striking correlation between atmospheric carbon dioxide concentration and global temperature. As both of these variables increase at significant rates, there are warnings of a planetary emergency that threatens to condemn coming generations to a catastrophically diminished future.*

In this chapter, you’ll learn to approach our climate crisis mathematically by creating formulas, called functions, that model data for average global temperature and carbon dioxide concentration over time. Understanding the concept of a function will give you a new perspective on many situations, ranging from climate change to using mathematics in a way that is similar to making a movie.

*Sources: Al Gore, An Inconvenient Truth, Rodale, 2006; Time, April 3, 2006; Rolling Stone, September 26, 2013

Here’s where you’ll find these applications:

Section 1.1: Graphs and Graphing Utilities

Section 1.1 Graphs and Graphing Utilities

Learning Objectives

What You’ll Learn

  1. 1 Plot points in the rectangular coordinate system.

  2. 2 Graph equations in the rectangular coordinate system.

  3. 3 Interpret information about a graphing utility’s viewing rectangle or table.

  4. 4 Use a graph to determine intercepts.

  5. 5 Interpret information given by graphs.

The beginning of the seventeenth century was a time of innovative ideas and enormous intellectual progress in Europe. English theatergoers enjoyed a succession of exciting new plays by Shakespeare. William Harvey proposed the radical notion that the heart was a pump for blood rather than the center of emotion. Galileo, with his newfangled invention called the telescope, supported the theory of Polish astronomer Copernicus that the Sun, not the Earth, was the center of the solar system. Monteverdi was writing the world’s first grand operas. French mathematicians Pascal and Fermat invented a new field of mathematics called probability theory.

Into this arena of intellectual electricity stepped French aristocrat René Descartes (1596–1650). Descartes (pronounced “day cart”), propelled by the creativity surrounding him, developed a new branch of mathematics that brought together algebra and geometry in a unified way—a way that visualized numbers as points on a graph, equations as geometric figures, and geometric figures as equations. This new branch of mathematics, called analytic geometry, established Descartes as one of the founders of modern thought and among the most original mathematicians and philosophers of any age. We begin this section by looking at Descartes’s deceptively simple idea, called the rectangular coordinate system or (in his honor) the Cartesian coordinate system.

Objective 1: Plot points in the rectangular coordinate system

Points and Ordered Pairs

  1. Objective 1 Plot points in the rectangular coordinate system.

Watch Video

Descartes used two number lines that intersect at right angles at their zero points, as shown in Figure 1.1. The horizontal number line is the x-axis. The vertical number line is the y-axis. The point of intersection of these axes is their zero points, called the origin. Positive numbers are shown to the right and above the origin. Negative numbers are shown to the left and below the origin. The axes divide the plane into four quarters, called quadrants. The points located on the axes are not in any quadrant.

Figure 1.1 The rectangular coordinate system

A diagram shows the rectangular coordinate system.
Figure 1.1 Full Alternative Text

Each point in the rectangular coordinate system corresponds to an ordered pair of real numbers, (x, y). Examples of such pairs are (5, 3) and (3, 5). The first number in each pair, called the x-coordinate, denotes the distance and direction from the origin along the x-axis. The second number in each pair, called the y-coordinate, denotes vertical distance and direction along a line parallel to the y-axis or along the y-axis itself.

Figure 1.2 shows how we plot, or locate, the points corresponding to the ordered pairs (5, 3) and (3, 5). We plot (5, 3) by going 5 units from 0 to the left along the x-axis. Then we go 3 units up parallel to the y-axis. We plot (3, 5) by going 3 units from 0 to the right along the x-axis and 5 units down parallel to the y-axis. The phrase “the points corresponding to the ordered pairs (5, 3) and (3, 5)” is often abbreviated as “the points (5, 3) and (3, 5).”

Figure 1.2 Plotting (5, 3) and (3, 5)

A graph plots two points in the rectangular coordinate system.
Figure 1.2 Full Alternative Text

Example 1 Plotting Points in the Rectangular Coordinate System

Plot the points: A(3, 5), B(2, 4), C(5, 0), D(5, 3), E(0, 4), and F(0, 0).

Solution

See Figure 1.3. We move from the origin and plot the points in the following way:

F left parenthesis 0 comma 0 right parenthesis. 0 units right or left comma 0 units up or down.

Figure 1.3 Plotting points

A diagram plots six points in the rectangular coordinate system.

Check Point 1

  • Plot the points: A(2, 4), B(4, 2), C(3, 0), and D(0, 3).

Objective 2: Graph equations in the rectangular coordinate system

Graphs of Equations

  1. Objective 2 Graph equations in the rectangular coordinate system.

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A relationship between two quantities can be expressed as an equation in two variables, such as

y=4x2.

A solution of an equation in two variables, x and y, is an ordered pair of real numbers with the following property: When the x-coordinate is substituted for x and the y-coordinate is substituted for y in the equation, we obtain a true statement. For example, consider the equation y=4x2 and the ordered pair (3, 5). When 3 is substituted for x and 5 is substituted for y, we obtain the statement 5=432, or 5=49, or 5=5. Because this statement is true, the ordered pair (3, 5) is a solution of the equation y=4x2. We also say that (3, 5) satisfies the equation.

We can generate as many ordered-pair solutions as desired to y=4x2 by substituting numbers for x and then finding the corresponding values for y. For example, suppose we let x=3:

An illustration shows a table to find the ordered pair from an equation.
1.1-3 Full Alternative Text

The graph of an equation in two variables is the set of all points whose coordinates satisfy the equation. One method for graphing such equations is the point-plotting method. First, we find several ordered pairs that are solutions of the equation. Next, we plot these ordered pairs as points in the rectangular coordinate system. Finally, we connect the points with a smooth curve or line. This often gives us a picture of all ordered pairs that satisfy the equation.

Example 2 Graphing an Equation Using the Point-Plotting Method

Graph y=4x2. Select integers for x, starting with 3 and ending with 3.

Solution

For each value of x, we find the corresponding value for y.

An illustration shows finding the ordered pair from an equation.

Now we plot the seven points and join them with a smooth curve, as shown in Figure 1.4. The graph of y=4x2 is a curve where the part of the graph to the right of the y-axis is a reflection of the part to the left of it and vice versa. The arrows on the left and the right of the curve indicate that it extends indefinitely in both directions.

Figure 1.4 The graph of y=4x2

A rectangular coordinate system shows a graph of downward opening parabola with indefinite ends.

Check Point 2

  • Graph y=4x. Select integers for x, starting with 3 and ending with 3.

Example 3 Graphing an Equation Using the Point-Plotting Method

Graph y=|x|. Select integers for x, starting with 3 and ending with 3.

Solution

For each value of x, we find the corresponding value for y.

x y=|x| Ordered Pair (x, y)

3

    y=|−3|=3

(3, 3)

2

    y=|−2|=2

(2, 2)

1

    y=|−1|=1

(1, 1)

  0

y=|0|=0

(0, 0)

  1

y=|1|=1

(1, 1)

  2

y=|2|=2

(2, 2)

  3

y=|3|=3

(3, 3)

We plot the points and connect them, resulting in the graph shown in Figure 1.5. The graph is V-shaped and centered at the origin. For every point (x, y) on the graph, the point (x, y) is also on the graph. This shows that the absolute value of a positive number is the same as the absolute value of its opposite.

Figure 1.5 The graph of y=|x|

A rectangular coordinate system shows a graph of two lines intersecting at the origin.

Check Point 3

  • Graph y=|x+1|. Select integers for x, starting with 4 and ending with 2.

Objective 2: Graph equations in the rectangular coordinate system

Graphs of Equations

  1. Objective 2 Graph equations in the rectangular coordinate system.

Watch Video

A relationship between two quantities can be expressed as an equation in two variables, such as

y=4x2.

A solution of an equation in two variables, x and y, is an ordered pair of real numbers with the following property: When the x-coordinate is substituted for x and the y-coordinate is substituted for y in the equation, we obtain a true statement. For example, consider the equation y=4x2 and the ordered pair (3, 5). When 3 is substituted for x and 5 is substituted for y, we obtain the statement 5=432, or 5=49, or 5=5. Because this statement is true, the ordered pair (3, 5) is a solution of the equation y=4x2. We also say that (3, 5) satisfies the equation.

We can generate as many ordered-pair solutions as desired to y=4x2 by substituting numbers for x and then finding the corresponding values for y. For example, suppose we let x=3:

An illustration shows a table to find the ordered pair from an equation.
1.1-3 Full Alternative Text

The graph of an equation in two variables is the set of all points whose coordinates satisfy the equation. One method for graphing such equations is the point-plotting method. First, we find several ordered pairs that are solutions of the equation. Next, we plot these ordered pairs as points in the rectangular coordinate system. Finally, we connect the points with a smooth curve or line. This often gives us a picture of all ordered pairs that satisfy the equation.

Example 2 Graphing an Equation Using the Point-Plotting Method

Graph y=4x2. Select integers for x, starting with 3 and ending with 3.

Solution

For each value of x, we find the corresponding value for y.

An illustration shows finding the ordered pair from an equation.

Now we plot the seven points and join them with a smooth curve, as shown in Figure 1.4. The graph of y=4x2 is a curve where the part of the graph to the right of the y-axis is a reflection of the part to the left of it and vice versa. The arrows on the left and the right of the curve indicate that it extends indefinitely in both directions.

Figure 1.4 The graph of y=4x2

A rectangular coordinate system shows a graph of downward opening parabola with indefinite ends.

Check Point 2

  • Graph y=4x. Select integers for x, starting with 3 and ending with 3.

Example 3 Graphing an Equation Using the Point-Plotting Method

Graph y=|x|. Select integers for x, starting with 3 and ending with 3.

Solution

For each value of x, we find the corresponding value for y.

x y=|x| Ordered Pair (x, y)

3

    y=|−3|=3

(3, 3)

2

    y=|−2|=2

(2, 2)

1

    y=|−1|=1

(1, 1)

  0

y=|0|=0

(0, 0)

  1

y=|1|=1

(1, 1)

  2

y=|2|=2

(2, 2)

  3

y=|3|=3

(3, 3)

We plot the points and connect them, resulting in the graph shown in Figure 1.5. The graph is V-shaped and centered at the origin. For every point (x, y) on the graph, the point (x, y) is also on the graph. This shows that the absolute value of a positive number is the same as the absolute value of its opposite.

Figure 1.5 The graph of y=|x|

A rectangular coordinate system shows a graph of two lines intersecting at the origin.

Check Point 3

  • Graph y=|x+1|. Select integers for x, starting with 4 and ending with 2.

Objective 3: Interpret information about a graphing utility’s viewing rectangle or table

Graphing Equations and Creating Tables Using a Graphing Utility

  1. Objective 3 Interpret information about a graphing utility’s viewing rectangle or table.

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Graphing calculators and graphing software packages for computers are referred to as graphing utilities or graphers. A graphing utility is a powerful tool that quickly generates the graph of an equation in two variables. Figures 1.6(a) and 1.6(b) show two such graphs for the equations in Examples 2 and 3.

Figure 1.6(a) The graph of y=4x2

A graphing calculator screen plots a graph of the downward opening parabola.

Figure 1.6(a) Full Alternative Text

Figure 1.6(b) The graph of y=|x|

A graphing calculator screen shows a graph of the two lines intersecting at the origin.

Figure 1.6(b) Full Alternative Text

What differences do you notice between these graphs and the graphs that we drew by hand? They do seem a bit “jittery.” Arrows do not appear on the left and right ends of the graphs. Furthermore, numbers are not given along the axes. For both graphs in Figure 1.6, the x-axis extends from 10 to 10 and the y-axis also extends from 10 to 10. The distance represented by each consecutive tick mark is one unit. We say that the viewing rectangle, or the viewing window, is [10, 10, 1] by [10, 10, 1].

A chart depicts the measurements of a viewing window on a graphing calculator screen.
1.1-6 Full Alternative Text

To graph an equation in x and y using a graphing utility, enter the equation and specify the size of the viewing rectangle. The size of the viewing rectangle sets minimum and maximum values for both the x- and y-axes. Enter these values, as well as the values representing the distances between consecutive tick marks, on the respective axes. The [10, 10, 1] by [10, 10, 1] viewing rectangle used in Figure 1.6 is called the standard viewing rectangle.

Example 4 Understanding the Viewing Rectangle

What is the meaning of a [2, 3, 0.5] by [10, 20, 5] viewing rectangle?

Solution

We begin with [2, 3, 0.5], which describes the x-axis. The minimum x-value is 2 and the maximum x-value is 3. The distance between consecutive tick marks is 0.5.

Next, consider [10, 20, 5], which describes the y-axis. The minimum y-value is 10 and the maximum y-value is 20. The distance between consecutive tick marks is 5.

Figure 1.7 illustrates a [2, 3, 0.5] by [10, 20, 5] viewing rectangle. To make things clearer, we’ve placed numbers by each tick mark. These numbers do not appear on the axes when you use a graphing utility to graph an equation.

Figure 1.7 A [2, 3, 0.5] by [10, 20, 5] viewing rectangle

An illustration shows the viewing window of a graphing calculator with x axis ranging from negative 2 to 3 in increments of 0.5 and y axis ranging from negative 10 to 20 in increments of 5.

Check Point 4

  • What is the meaning of a [100, 100, 50] by [100, 100, 10] viewing rectangle? Create a figure like the one in Figure 1.7 that illustrates this viewing rectangle.

On many graphing utilities, the display screen is five-eighths as high as it is wide. By using a square setting, you can equally space the x and y tick marks. (This does not occur in the standard viewing rectangle.) Graphing utilities can also zoom in and zoom out. When you zoom in, you see a smaller portion of the graph, but you do so in greater detail. When you zoom out, you see a larger portion of the graph. Thus, zooming out may help you to develop a better understanding of the overall character of the graph. With practice, you will become more comfortable with graphing equations in two variables using your graphing utility. You will also develop a better sense of the size of the viewing rectangle that will reveal needed information about a particular graph.

Graphing utilities can also be used to create tables showing solutions of equations in two variables. Use the Table Setup function to choose the starting value of x and to input the increment, or change, between the consecutive x-values. The corresponding y-values are calculated based on the equation(s) in two variables in the Y= screen. In Figure 1.8, we used a TI-84 Plus C to create a table for y=4x2 and y=|x|, the equations in Examples 2 and 3.

Figure 1.8 Creating a table for y1=4x2 and y2=|x|

The image shows a set of three screenshots of a graphing calculator screen depicts creating a table for two equations.

Figure 1.8 Full Alternative Text
Objective 4: Use a graph to determine intercepts

Intercepts

  1. Objective 4 Use a graph to determine intercepts.

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An x-intercept of a graph is the x-coordinate of a point where the graph intersects the x-axis. For example, look at the graph of y=4x2 in Figure 1.9 at the top of the next page. The graph crosses the x-axis at (2, 0) and (2, 0). Thus, the x-intercepts are 2 and 2. The y-coordinate corresponding to an x-intercept is always zero.

Figure 1.9 Intercepts of y=4x2

A rectangular coordinate system shows a graph of downward opening parabola with indefinite ends.

Figure 1.9 Full Alternative Text

A y-intercept of a graph is the y-coordinate of a point where the graph intersects the y-axis. The graph of y=4x2 in Figure 1.9 shows that the graph crosses the y-axis at (0, 4). Thus, the y-intercept is 4. The x-coordinate corresponding to a y-intercept is always zero.

Example 5 Identifying Intercepts

Identify the x- and y-intercepts.

  1.  

    A graph plots a line that slopes upward from the third quadrant to the first quadrant passing through the second quadrant. The line passes through the points (negative 1, 0) and (0, 2).
  2.  

    A graph plots a vertical line parallel to the y axis passing through the point (3, 0) on the x axis from the first quadrant to the fourth quadrant.
  3.  

    A graph plots a line that falls from the estimated points (negative 1, 2), (1, negative 2) passing through the origin (0, 0).
  4.  

    A rectangular coordinate system plots a graph of two lines intersecting at (0, negative 3).

Solution

  1. The graph crosses the x-axis at (1, 0). Thus, the x-intercept is 1. The graph crosses the y-axis at (0, 2). Thus, the y-intercept is 2.

  2. The graph crosses the x-axis at (3, 0), so the x-intercept is 3. This vertical line does not cross the y-axis. Thus, there is no y-intercept.

  3. This graph crosses the x- and y-axes at the same point, the origin. Because the graph crosses both axes at (0, 0), the x-intercept is 0 and the y-intercept is 0.

  4. The graph crosses the x-axis at (2,0) and (2,0). Thus, the x-intercepts are 2 and 2. The graph crosses the y-axis at (0, 3). Thus, the y-intercept is 3.

Check Point 5

  • Identify the x- and y-intercepts.

    1.  

      A graph plots a line that is rising through the points (negative 3, 0) and (0, 5).
    2.  

      A graph plots a horizontal line parallel to the x axis passing through the point (0, 4).
    3.  

      A graph plots a line rising through the origin. The line passes through the estimated points (negative 3, negative 3), (0, 0), and (3, 3).
    4.  

      A rectangular coordinate system plots a graph of two lines intersecting at point (0, 3). the first line falls from points (0, 3), (1, 0), and (3, negative 3). The second line rises from points (negative 3, negative 3), (negative 1, 0), (0, 3).

Figure 1.10 illustrates that a graph may have no intercepts or several intercepts.

Figure 1.10

An illustration shows graphs with no or multiple intercepts.
Figure 1.10 Full Alternative Text
An illustration shows graphs with no or multiple intercepts.
1.1-15 Full Alternative Text
An illustration shows graphs with no or multiple intercepts. The circle in the first quadrant and the text reads, no intercepts.
An illustration shows graphs with no or multiple intercepts.
1.1-15 Full Alternative Text
An illustration shows graphs with no or multiple intercepts. An upward opening parabola in the first and second quadrants with indefinite ends with its vertex at the origin. The text reads, the same x intercept and y intercept.
Objective 4: Use a graph to determine intercepts

Intercepts

  1. Objective 4 Use a graph to determine intercepts.

Watch Video

An x-intercept of a graph is the x-coordinate of a point where the graph intersects the x-axis. For example, look at the graph of y=4x2 in Figure 1.9 at the top of the next page. The graph crosses the x-axis at (2, 0) and (2, 0). Thus, the x-intercepts are 2 and 2. The y-coordinate corresponding to an x-intercept is always zero.

Figure 1.9 Intercepts of y=4x2

A rectangular coordinate system shows a graph of downward opening parabola with indefinite ends.

Figure 1.9 Full Alternative Text

A y-intercept of a graph is the y-coordinate of a point where the graph intersects the y-axis. The graph of y=4x2 in Figure 1.9 shows that the graph crosses the y-axis at (0, 4). Thus, the y-intercept is 4. The x-coordinate corresponding to a y-intercept is always zero.

Example 5 Identifying Intercepts

Identify the x- and y-intercepts.

  1.  

    A graph plots a line that slopes upward from the third quadrant to the first quadrant passing through the second quadrant. The line passes through the points (negative 1, 0) and (0, 2).
  2.  

    A graph plots a vertical line parallel to the y axis passing through the point (3, 0) on the x axis from the first quadrant to the fourth quadrant.
  3.  

    A graph plots a line that falls from the estimated points (negative 1, 2), (1, negative 2) passing through the origin (0, 0).
  4.  

    A rectangular coordinate system plots a graph of two lines intersecting at (0, negative 3).

Solution

  1. The graph crosses the x-axis at (1, 0). Thus, the x-intercept is 1. The graph crosses the y-axis at (0, 2). Thus, the y-intercept is 2.

  2. The graph crosses the x-axis at (3, 0), so the x-intercept is 3. This vertical line does not cross the y-axis. Thus, there is no y-intercept.

  3. This graph crosses the x- and y-axes at the same point, the origin. Because the graph crosses both axes at (0, 0), the x-intercept is 0 and the y-intercept is 0.

  4. The graph crosses the x-axis at (2,0) and (2,0). Thus, the x-intercepts are 2 and 2. The graph crosses the y-axis at (0, 3). Thus, the y-intercept is 3.

Check Point 5

  • Identify the x- and y-intercepts.

    1.  

      A graph plots a line that is rising through the points (negative 3, 0) and (0, 5).
    2.  

      A graph plots a horizontal line parallel to the x axis passing through the point (0, 4).
    3.  

      A graph plots a line rising through the origin. The line passes through the estimated points (negative 3, negative 3), (0, 0), and (3, 3).
    4.  

      A rectangular coordinate system plots a graph of two lines intersecting at point (0, 3). the first line falls from points (0, 3), (1, 0), and (3, negative 3). The second line rises from points (negative 3, negative 3), (negative 1, 0), (0, 3).

Figure 1.10 illustrates that a graph may have no intercepts or several intercepts.

Figure 1.10

An illustration shows graphs with no or multiple intercepts.
Figure 1.10 Full Alternative Text
An illustration shows graphs with no or multiple intercepts.
1.1-15 Full Alternative Text
An illustration shows graphs with no or multiple intercepts. The circle in the first quadrant and the text reads, no intercepts.
An illustration shows graphs with no or multiple intercepts.
1.1-15 Full Alternative Text
An illustration shows graphs with no or multiple intercepts. An upward opening parabola in the first and second quadrants with indefinite ends with its vertex at the origin. The text reads, the same x intercept and y intercept.
Objective 5: Interpret information given by graphs

Interpreting Information Given by Graphs

  1. Objective 5 Interpret information given by graphs.

Watch Video

Line graphs are often used to illustrate trends over time. Some measure of time, such as months or years, frequently appears on the horizontal axis. Amounts are generally listed on the vertical axis. Points are drawn to represent the given information. The graph is formed by connecting the points with line segments.

A line graph displays information in the first quadrant of a rectangular coordinate system. By identifying points on line graphs and their coordinates, you can interpret specific information given by the graph.

Example 6 Educational Attainment and the Probability of Divorce

Many factors affect whether a marriage will last. Data show that divorce rates are higher for those who marry younger, but education may also play a role in the longevity of a marriage. The line graphs in Figure 1.11 show the percentages of marriages ending in divorce after 5, 10, and 15 years of marriage for two levels of educational attainment.

Figure 1.11

A line graph depicts the probability of divorce by educational attainment.

Source: U.S. Bureau of Labor Statistics

Here are two mathematical models that approximate the data displayed by the line graphs:

d = 1.8n + 14. d = 1.3n + 6. A dialogue box pointing at the equation d = 1.8n + 14 reads, high school comma no college. A dialogue box pointing at the equation d = 1.3n + 6 reads, bachelor prime s degree or higher

In each model, the variable n is the number of years after marriage and the variable d is the percentage of marriages ending in divorce.

  1. Use the appropriate formula to determine the percentage of marriages ending in divorce after 10 years when the educational attainment is bachelor’s degree or higher.

  2. Use the appropriate line graph in Figure 1.11 to determine the percentage of marriages ending in divorce after 10 years when the educational attainment is bachelor’s degree or higher.

  3. Does the value given by the mathematical model underestimate or overestimate the actual percentage of marriages ending in divorce after 10 years as shown by the graph? By how much?

Solution

  1. Because the educational attainment is bachelor’s degree or higher, we use the formula on the right, d=1.3n+6. To find the percentage of marriages ending in divorce after 10 years, we substitute 10 for n and evaluate the formula.

    d=1.3n+6This is one of the two givenmathematical models.d=1.3(10)+6Replace n with 10.d=13+6Multiply: 1.3(10)=13.d=19Add.

    The model indicates that 19% of marriages end in divorce after 10 years for those with a bachelor’s degree or higher.

  2. Now let’s use the line graph that shows the percentage of marriages ending in divorce for a bachelor’s degree or higher. The graph is shown again in Figure 1.12. To find the percentage of marriages ending in divorce after 10 years:

    • Locate 10 on the horizontal axis and locate the point above 10.

    • Read across to the corresponding percent on the vertical axis.

    Figure 1.12

    A line graph depicts the probability of divorce for the educational attainment of bachelor prime s degree or higher.

    The actual data displayed by the graph indicate that 20% of these marriages end in divorce after 10 years.

  3. The value obtained by evaluating the mathematical model, 19%, is close to, but slightly less than, the actual percentage of divorces, 20%. The difference between these percents is 20%19%, or 1%. The value given by the mathematical model, 19%, underestimates the actual percent, 20%, by only 1, providing a fairly accurate description of the data.

    The data presented in Example 6 indicate longer-lasting marriages for those with more education. Does this mean that a college education prepares you for marriage? Not necessarily. Although the academics and social interactions in college may broaden a person’s perspective of relationships, there are underlying issues at play here. One such issue is finances: More education typically means higher earnings, which may lead to greater financial stability. Financial stress in a marriage is a major contributor to divorce.

    Throughout this text, you will encounter models based on real-world data. To keep our models manageable, we limit the number of factors under consideration. However, we want you to be aware of the existence of underlying issues that may offer an alternative explanation for the trends observed in our numerous graphs and models.

    Our goal is that you acquire a greater appreciation of mathematical models and their applications both in other academic disciplines and in real life, but we also want you to be aware of the limitations of these models.

Check Point 6

    1. Use the appropriate formula from Example 6 to determine the percentage of marriages ending in divorce after 15 years for high school graduates with no college.

    2. Use the appropriate line graph in Figure 1.11 to determine the percentage of marriages ending in divorce after 15 years for high school graduates with no college.

    3. Does the value given by the mathematical model underestimate or overestimate the actual percentage of marriages ending in divorce after 15 years as shown by the graph? By how much?

Objective 5: Interpret information given by graphs

Interpreting Information Given by Graphs

  1. Objective 5 Interpret information given by graphs.

Watch Video

Line graphs are often used to illustrate trends over time. Some measure of time, such as months or years, frequently appears on the horizontal axis. Amounts are generally listed on the vertical axis. Points are drawn to represent the given information. The graph is formed by connecting the points with line segments.

A line graph displays information in the first quadrant of a rectangular coordinate system. By identifying points on line graphs and their coordinates, you can interpret specific information given by the graph.

Example 6 Educational Attainment and the Probability of Divorce

Many factors affect whether a marriage will last. Data show that divorce rates are higher for those who marry younger, but education may also play a role in the longevity of a marriage. The line graphs in Figure 1.11 show the percentages of marriages ending in divorce after 5, 10, and 15 years of marriage for two levels of educational attainment.

Figure 1.11

A line graph depicts the probability of divorce by educational attainment.

Source: U.S. Bureau of Labor Statistics

Here are two mathematical models that approximate the data displayed by the line graphs:

d = 1.8n + 14. d = 1.3n + 6. A dialogue box pointing at the equation d = 1.8n + 14 reads, high school comma no college. A dialogue box pointing at the equation d = 1.3n + 6 reads, bachelor prime s degree or higher

In each model, the variable n is the number of years after marriage and the variable d is the percentage of marriages ending in divorce.

  1. Use the appropriate formula to determine the percentage of marriages ending in divorce after 10 years when the educational attainment is bachelor’s degree or higher.

  2. Use the appropriate line graph in Figure 1.11 to determine the percentage of marriages ending in divorce after 10 years when the educational attainment is bachelor’s degree or higher.

  3. Does the value given by the mathematical model underestimate or overestimate the actual percentage of marriages ending in divorce after 10 years as shown by the graph? By how much?

Solution

  1. Because the educational attainment is bachelor’s degree or higher, we use the formula on the right, d=1.3n+6. To find the percentage of marriages ending in divorce after 10 years, we substitute 10 for n and evaluate the formula.

    d=1.3n+6This is one of the two givenmathematical models.d=1.3(10)+6Replace n with 10.d=13+6Multiply: 1.3(10)=13.d=19Add.

    The model indicates that 19% of marriages end in divorce after 10 years for those with a bachelor’s degree or higher.

  2. Now let’s use the line graph that shows the percentage of marriages ending in divorce for a bachelor’s degree or higher. The graph is shown again in Figure 1.12. To find the percentage of marriages ending in divorce after 10 years:

    • Locate 10 on the horizontal axis and locate the point above 10.

    • Read across to the corresponding percent on the vertical axis.

    Figure 1.12

    A line graph depicts the probability of divorce for the educational attainment of bachelor prime s degree or higher.

    The actual data displayed by the graph indicate that 20% of these marriages end in divorce after 10 years.

  3. The value obtained by evaluating the mathematical model, 19%, is close to, but slightly less than, the actual percentage of divorces, 20%. The difference between these percents is 20%19%, or 1%. The value given by the mathematical model, 19%, underestimates the actual percent, 20%, by only 1, providing a fairly accurate description of the data.

    The data presented in Example 6 indicate longer-lasting marriages for those with more education. Does this mean that a college education prepares you for marriage? Not necessarily. Although the academics and social interactions in college may broaden a person’s perspective of relationships, there are underlying issues at play here. One such issue is finances: More education typically means higher earnings, which may lead to greater financial stability. Financial stress in a marriage is a major contributor to divorce.

    Throughout this text, you will encounter models based on real-world data. To keep our models manageable, we limit the number of factors under consideration. However, we want you to be aware of the existence of underlying issues that may offer an alternative explanation for the trends observed in our numerous graphs and models.

    Our goal is that you acquire a greater appreciation of mathematical models and their applications both in other academic disciplines and in real life, but we also want you to be aware of the limitations of these models.

Check Point 6

    1. Use the appropriate formula from Example 6 to determine the percentage of marriages ending in divorce after 15 years for high school graduates with no college.

    2. Use the appropriate line graph in Figure 1.11 to determine the percentage of marriages ending in divorce after 15 years for high school graduates with no college.

    3. Does the value given by the mathematical model underestimate or overestimate the actual percentage of marriages ending in divorce after 15 years as shown by the graph? By how much?

1.1: Exercise Set

1.1 Exercise Set

Practice Exercises

In Exercises 112, plot the given point in a rectangular coordinate system.

  1. 1. (1, 4)

  2. 2. (2, 5)

  3. 3. (2, 3)

  4. 4. (1, 4)

  5. 5. (3, 5)

  6. 6. (4, 2)

  7. 7. (4, 1)

  8. 8. (3, 2)

  9. 9. (4, 0)

  10. 10. (0, 3)

  11. 11. (72, 32)

  12. 12. (52, 32)

Graph each equation in Exercises 1328. Let x=3, 2, 1, 0, 1, 2, and 3.

  1. 13. y=x22

  2. 14. y=x2+2

  3. 15. y=x2

  4. 16. y=x+2

  5. 17. y=2x+1

  6. 18. y=2x4

  7. 19. y=12 x

  8. 20. y=12 x+2

  9. 21. y=2|x|

  10. 22. y=2|x|

  11. 23. y=|x|+1

  12. 24. y=|x|1

  13. 25. y=9x2

  14. 26. y=x2

  15. 27. y=x3

  16. 28. y=x31

In Exercises 2932, match the viewing rectangle with the correct figure. Then label the tick marks in the figure to illustrate this viewing rectangle.

  1. 29. [5, 5, 1] by [5, 5, 1]

  2. 30. [10, 10, 2] by [4, 4, 2]

  3. 31. [20, 80, 10] by [30, 70, 10]

  4. 32. [40, 40, 20] by [1000, 1000, 100]

    1.  

      A graphing calculator screen shows the reviewing rectangle as the x axis with 5 range marks and the y axis with 21 marks. The two axes intersect at the middle mark of each axis.
    2.  

      A graphing calculator screen shows the reviewing rectangle as the x axis with 11 range marks and the y axis with range marks. The two axes intersect at the third mark from the left of x axis and the fourth mark from the bottom of the y axis.
    3.  

      A graphing calculator screenshows the reviewing rectangle as the x axis with 11 range marks and the y axis with 11 range marks. The two axes intersect at the middle mark of each axis.
    4.  

      A graphing calculator screenshows the reviewing rectangle as the x axis with 11 range marks and the y axis with 5 range marks. The two axes intersect at the middle mark of each.

The table of values was generated by a graphing utility with a TABLE feature. Use the table to solve Exercises 3340.

The image depicts the graphing calculator screentable.
  1. 33. Which equation corresponds to Y2 in the table?

    1. y2=x+8

    2. y2=x2

    3. y2=2x

    4. y2=12x

  2. 34. Which equation corresponds to Y1 in the table?

    1. y1=3x

    2. y1=x2

    3. y1=x2

    4. y1=2x

  3. 35. Does the graph of Y2 pass through the origin?

  4. 36. Does the graph of Y1 pass through the origin?

  5. 37. At which point does the graph of Y2 cross the x-axis?

  6. 38. At which point does the graph of Y2 cross the y-axis?

  7. 39. At which points do the graphs of Y1 and Y2 intersect?

  8. 40. For which values of x is Y1=Y2?

In Exercises 4146, use the graph to a. determine the x-intercepts, if any; b. determine the y-intercepts, if any. For each graph, tick marks along the axes represent one unit each.

  1. 41.

    A graph plots a line that rises from the third quadrant to the first quadrant passing through the fourth quadrant. The line intersects the y axis at the fourth range mark below the origin and x axis at the second range mark right to the origin.
  2. 42.

    A graph plots a line that slopes upward from the fourth quadrant to the second quadrant passing through the first quadrant. The line intersects x axis at the first tick mark right to the origin and the y axis at the second tick mark above the origin.

  3. 43.

    A rectangular coordinate system plots a graph of an N shaped curve.
  4. 44.

    A rectangular coordinate system plots a graph of a W shaped curve.
  5. 45.

    A graph plots a left opening parabola with indefinite ends in second and third quadrants, vertex at (1, 0).
  6. 46.

    A graph plots an upward opening parabola with indefinite ends, vertex (0, 2).

Practice PLUS

In Exercises 4750, write each English sentence as an equation in two variables. Then graph the equation.

  1. 47. The y-value is four more than twice the x-value.

  2. 48. The y-value is the difference between four and twice the x-value.

  3. 49. The y-value is three decreased by the square of the x-value.

  4. 50. The y-value is two more than the square of the x-value.

In Exercises 5154, graph each equation.

  1. 51. y=5 (Let x=3, 2, 1, 0, 1, 2, and 3.)

  2. 52. y=1 (Let x=3, 2, 1, 0, 1, 2, and 3.)

  3. 53. y=1x (Let x=2, 1, 12, 13, 13, 12, 1, and 2.)

  4. 54. y= 1x (Let x=2, 1, 12, 13, 13, 12, 1, and 2.)

Application Exercises

The graphs show the percentage of high school seniors who had ever used alcohol or marijuana.

A line graph depicts the alcohol and marijuana use by United States high school seniors.

Source: University of Michigan Institute for Social Research

The data can be described by the following mathematical models:

Percentage of seniors using alcohol, A equals negative 0.9 n + 88. Percentage of seniors using marijuana, M equals 0.1 lower n + 43. n is the number of years after 19 90. A dialogue box pointing both the equation reads, number of year 19 90.

Use this information to solve Exercises 5556.

  1. 55.

    1. Use the appropriate line graph to estimate the percentage of seniors who had ever used marijuana in 2010.

    2. Use the appropriate formula to determine the percentage of seniors who had ever used marijuana in 2010. How does this compare with your estimate in part (a)?

    3. Use the appropriate line graph to estimate the percentage of seniors who had ever used alcohol in 2010.

    4. Use the appropriate formula to determine the percentage of seniors who had ever used alcohol in 2010. How does this compare with your estimate in part (c)?

    5. For the period from 1990 through 2019, in which year was marijuana use by seniors at a maximum? Estimate the percentage of seniors who had ever used marijuana in that year.

  2. 56.

    1. Use the appropriate line graph to estimate the percentage of seniors who had ever used alcohol in 2015.

    2. Use the appropriate formula to determine the percentage of seniors who had ever used alcohol in 2015. How does this compare with your estimate in part (a)?

    3. Use the appropriate line graph to estimate the percentage of seniors who had ever used marijuana in 2015.

    4. Use the appropriate formula to determine the percentage of seniors who had ever used marijuana in 2015. How does this compare with your estimate in part (c)?

    5. For the period from 1990 through 2019, in which year was alcohol use by seniors at a maximum? What percentage of seniors had ever used alcohol in that year?

Contrary to popular belief, older people do not need less sleep than younger adults. However, the line graphs show that they awaken more often during the night. The numerous awakenings are one reason why some elderly individuals report that sleep is less restful than it had been in the past. Use the line graphs to solve Exercises 5760.

A line graph depicts the average number of awakenings during the night by age and gender.

Source: Stephen Davis and Joseph Palladino, Psychology, 5th Edition, Prentice Hall, 2007

  1. 57. At which age, estimated to the nearest year, do women have the least number of awakenings during the night? What is the average number of awakenings at that age?

  2. 58. At which age do men have the greatest number of awakenings during the night? What is the average number of awakenings at that age?

  3. 59. Estimate, to the nearest tenth, the difference between the average number of awakenings during the night for 25-year-old men and 25-year-old women.

  4. 60. Estimate, to the nearest tenth, the difference between the average number of awakenings during the night for 18-year-old men and 18-year-old women.

Explaining the Concepts

  1. 61. What is the rectangular coordinate system?

  2. 62. Explain how to plot a point in the rectangular coordinate system. Give an example with your explanation.

  3. 63. Explain why (5, 2) and (2, 5) do not represent the same point.

  4. 64. Explain how to graph an equation in the rectangular coordinate system.

  5. 65. What does a [20, 2, 1] by [4, 5, 0.5] viewing rectangle mean?

Technology Exercise

  1. 66. Use a graphing utility to verify each of your hand-drawn graphs in Exercises 1328. Experiment with the settings for the viewing rectangle to make the graph displayed by the graphing utility resemble your hand-drawn graph as much as possible.

Critical Thinking Exercises

MAKE SENSE? In Exercises 6770, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 67. The rectangular coordinate system provides a geometric picture of what an equation in two variables looks like.

  2. 68. There is something wrong with my graphing utility because it is not displaying numbers along the x- and y-axes.

  3. 69. I used the ordered pairs (2, 2), (0, 0), and (2, 2) to graph a straight line.

  4. 70. I used the ordered pairs

         (time of day, calories that I burned)

    to obtain a graph that is a horizontal line.

In Exercises 7174, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 71. If the product of a point’s coordinates is positive, the point must be in quadrant I.

  2. 72. If a point is on the x-axis, it is neither up nor down, so x=0.

  3. 73. If a point is on the y-axis, its x-coordinate must be 0.

  4. 74. The ordered pair (2, 5) satisfies 3y2x=4.

In Exercises 7578, list the quadrant or quadrants satisfying each condition.

  1. 75. xy>0

  2. 76. yx<0

  3. 77. x3>0 and y3<0

  4. 78. x3<0 and y3>0

In Exercises 7982, match the story with the correct figure. The figures are labeled (a), (b), (c), and (d).

  1. 79. As the blizzard got worse, the snow fell harder and harder.

  2. 80. The snow fell more and more softly.

  3. 81. It snowed hard, but then it stopped. After a short time, the snow started falling softly.

  4. 82. It snowed softly, and then it stopped. After a short time, the snow started falling hard.

    1.  

      The graph shows the amount of snowfall versus time, graph shows a concave up curve that slopes upward from origin to the upper right corner.
    2.  

      The graph shows the amount of snowfall versus the time graph shows a line that slopes upward from the origin becomes parallel to the horizontal axis and then again slopes upwards to the upper right corner.
    3.  

      The graph shows the amount of snowfall versus the time, graph shows a line that slopes upward from the origin, becomes parallel to the horizontal axis, and then slopes to the upper right corner.
    4.  

      The graph shows the amount of snowfall versus the time graph shows a concave down curve that slopes upward from origin to the upper right corner.

In Exercises 8387, select the graph that best illustrates each story.

  1. 83. An airplane flew from Miami to San Francisco. [Graphs (c) and (d) are at the top of the next page.]

    1.  

      The graph shows planes' height versus seconds after takeoff graph shows a line that slopes upward from the origin to the upper right corner.
    2.  

      The graph shows planes height versus seconds after takeoff graph shows a line that slopes upward from the origin becomes parallel to the horizontal axis and then slopes downward to meet the horizontal axis near the right end.

    3.  

      The graph shows plane's height versus seconds after takeoff graph shows a line that slopes downward from a point on the vertical axis near the top and then becomes parallel to the horizontal axis in the middle.
    4.  

      The graph shows planes' height versus seconds after takeoff graph shows a line that first moves parallel to the horizontal axis from a point on the vertical axis and then moves vertically downward to meet the horizontal axis near the right end.
  2. 84. At noon, you begin to breathe in.

    1.  

      The graph shows the volume of air in the lungs versus the time afternoon. The graph shows a curve that moves to the right from a vertical axis, first forming a concave down shape and then forming a concave up shape.
    2.  

      The graph shows the volume of air in lungs versus the time afternoon graph shows a curve that moves to the right from a point on the vertical axis, first forming a concave up shape and then forming a concave down shape.
    3.  

      The graph shows the volume of air in the lungs versus the time afternoon graph shows a line that moves parallel to the horizontal axis to the right from a point on the vertical axis.
    4.  

      The graph shows that the volume of air in the lungs versus the afternoon graph shows a line that slopes upward from a point on the vertical axis to the upper right corner.
  3. 85. Measurements are taken of a person’s height from birth to age 100.

    1.  

      The graph shows height versus age graph shows a concave down curve that slopes upward from a point just above the origin.
    2.  

      The graph shows height versus age graph shows a curve that slopes upward from a point just above the origin, first as a concave down curve and then as a concave up curve.
    3.  

      The graph shows height versus age graph shows a concave down curve that moves to the right from a point just above the origin.
    4.  

      The graph shows height versus age graph shows a line that slopes upward from a point just above the origin. The line then becomes parallel to the horizontal axis.
  4. 86. You begin your bike ride by riding down a hill. Then you ride up another hill. Finally, you ride along a level surface before coming to a stop. [Graphs (c) and (d) are at the top of the next column.]

    1.  

      The graph shows speed versus time graph shows a curve that slopes downward from a point on the vertical axis near the top.
    2.  

      The graph shows speed versus time graph shows a curve that slopes upward from the origin as a concave up curve becomes parallel to the horizontal axis and then slopes down to meet the horizontal axis.
    3.  

      The graph shows speed versus time graph plots a curve that slopes upward from the origin as a concave down curve and then slopes downward to meet the horizontal axis near the right end.
    4.  

      The graph shows speed versus time graph plots a curve that slopes downward from a point on the vertical axis near the top as a concave up curve and then becomes parallel to the horizontal axis near the right end.
  5. 87. In Example 6, we used the formula d=1.3n+6 to model the percentage of marriages ending in divorce, d, after n years of marriage for an educational attainment of bachelor’s degree or higher. We can also model the data with the formula

    d=8n6.

    Using a calculator, evaluate each formula for n=5, 10, and 15. Round to the nearest tenth, where necessary. Which model appears to give the better estimates of the percentages shown in Figure 1.11?

Group Exercise

  1. 88. The group should identify three free online graphing calculators. Each group member should graph five equations from this exercise set using each of the three graphing calculators. Then, as a group, discuss what you like or don’t like about the calculators. Based on this discussion, make a list of criteria that you would use in choosing a graphing calculator. Which, if any, of the three graphing calculators would you recommend to fellow students?

Preview Exercises

Exercises 8991 will help you prepare for the material covered in the next section.

  1. 89. Here are two sets of ordered pairs:

    set 1: {(1,5),(2,5)}set 2: {(5,1),(5,2)}.

    In which set is each x-coordinate paired with only one y-coordinate?

  2. 90. Graph y=2x and y=2x+4 in the same rectangular coordinate system. Select integers for x, starting with 2 and ending with 2.

  3. 91. Use the following graph to solve this exercise.

    A graph plots two lines as a V shaped curved intersecting at (0, 1).
    1. What is the y-coordinate when the x-coordinate is 2?

    2. What are the x-coordinates when the y-coordinate is 4?

    3. Describe the x-coordinates of all points on the graph.

    4. Describe the y-coordinates of all points on the graph.

1.1: Exercise Set

1.1 Exercise Set

Practice Exercises

In Exercises 112, plot the given point in a rectangular coordinate system.

  1. 1. (1, 4)

  2. 2. (2, 5)

  3. 3. (2, 3)

  4. 4. (1, 4)

  5. 5. (3, 5)

  6. 6. (4, 2)

  7. 7. (4, 1)

  8. 8. (3, 2)

  9. 9. (4, 0)

  10. 10. (0, 3)

  11. 11. (72, 32)

  12. 12. (52, 32)

Graph each equation in Exercises 1328. Let x=3, 2, 1, 0, 1, 2, and 3.

  1. 13. y=x22

  2. 14. y=x2+2

  3. 15. y=x2

  4. 16. y=x+2

  5. 17. y=2x+1

  6. 18. y=2x4

  7. 19. y=12 x

  8. 20. y=12 x+2

  9. 21. y=2|x|

  10. 22. y=2|x|

  11. 23. y=|x|+1

  12. 24. y=|x|1

  13. 25. y=9x2

  14. 26. y=x2

  15. 27. y=x3

  16. 28. y=x31

In Exercises 2932, match the viewing rectangle with the correct figure. Then label the tick marks in the figure to illustrate this viewing rectangle.

  1. 29. [5, 5, 1] by [5, 5, 1]

  2. 30. [10, 10, 2] by [4, 4, 2]

  3. 31. [20, 80, 10] by [30, 70, 10]

  4. 32. [40, 40, 20] by [1000, 1000, 100]

    1.  

      A graphing calculator screen shows the reviewing rectangle as the x axis with 5 range marks and the y axis with 21 marks. The two axes intersect at the middle mark of each axis.
    2.  

      A graphing calculator screen shows the reviewing rectangle as the x axis with 11 range marks and the y axis with range marks. The two axes intersect at the third mark from the left of x axis and the fourth mark from the bottom of the y axis.
    3.  

      A graphing calculator screenshows the reviewing rectangle as the x axis with 11 range marks and the y axis with 11 range marks. The two axes intersect at the middle mark of each axis.
    4.  

      A graphing calculator screenshows the reviewing rectangle as the x axis with 11 range marks and the y axis with 5 range marks. The two axes intersect at the middle mark of each.

The table of values was generated by a graphing utility with a TABLE feature. Use the table to solve Exercises 3340.

The image depicts the graphing calculator screentable.
  1. 33. Which equation corresponds to Y2 in the table?

    1. y2=x+8

    2. y2=x2

    3. y2=2x

    4. y2=12x

  2. 34. Which equation corresponds to Y1 in the table?

    1. y1=3x

    2. y1=x2

    3. y1=x2

    4. y1=2x

  3. 35. Does the graph of Y2 pass through the origin?

  4. 36. Does the graph of Y1 pass through the origin?

  5. 37. At which point does the graph of Y2 cross the x-axis?

  6. 38. At which point does the graph of Y2 cross the y-axis?

  7. 39. At which points do the graphs of Y1 and Y2 intersect?

  8. 40. For which values of x is Y1=Y2?

In Exercises 4146, use the graph to a. determine the x-intercepts, if any; b. determine the y-intercepts, if any. For each graph, tick marks along the axes represent one unit each.

  1. 41.

    A graph plots a line that rises from the third quadrant to the first quadrant passing through the fourth quadrant. The line intersects the y axis at the fourth range mark below the origin and x axis at the second range mark right to the origin.
  2. 42.

    A graph plots a line that slopes upward from the fourth quadrant to the second quadrant passing through the first quadrant. The line intersects x axis at the first tick mark right to the origin and the y axis at the second tick mark above the origin.

  3. 43.

    A rectangular coordinate system plots a graph of an N shaped curve.
  4. 44.

    A rectangular coordinate system plots a graph of a W shaped curve.
  5. 45.

    A graph plots a left opening parabola with indefinite ends in second and third quadrants, vertex at (1, 0).
  6. 46.

    A graph plots an upward opening parabola with indefinite ends, vertex (0, 2).

Practice PLUS

In Exercises 4750, write each English sentence as an equation in two variables. Then graph the equation.

  1. 47. The y-value is four more than twice the x-value.

  2. 48. The y-value is the difference between four and twice the x-value.

  3. 49. The y-value is three decreased by the square of the x-value.

  4. 50. The y-value is two more than the square of the x-value.

In Exercises 5154, graph each equation.

  1. 51. y=5 (Let x=3, 2, 1, 0, 1, 2, and 3.)

  2. 52. y=1 (Let x=3, 2, 1, 0, 1, 2, and 3.)

  3. 53. y=1x (Let x=2, 1, 12, 13, 13, 12, 1, and 2.)

  4. 54. y= 1x (Let x=2, 1, 12, 13, 13, 12, 1, and 2.)

Application Exercises

The graphs show the percentage of high school seniors who had ever used alcohol or marijuana.

A line graph depicts the alcohol and marijuana use by United States high school seniors.

Source: University of Michigan Institute for Social Research

The data can be described by the following mathematical models:

Percentage of seniors using alcohol, A equals negative 0.9 n + 88. Percentage of seniors using marijuana, M equals 0.1 lower n + 43. n is the number of years after 19 90. A dialogue box pointing both the equation reads, number of year 19 90.

Use this information to solve Exercises 5556.

  1. 55.

    1. Use the appropriate line graph to estimate the percentage of seniors who had ever used marijuana in 2010.

    2. Use the appropriate formula to determine the percentage of seniors who had ever used marijuana in 2010. How does this compare with your estimate in part (a)?

    3. Use the appropriate line graph to estimate the percentage of seniors who had ever used alcohol in 2010.

    4. Use the appropriate formula to determine the percentage of seniors who had ever used alcohol in 2010. How does this compare with your estimate in part (c)?

    5. For the period from 1990 through 2019, in which year was marijuana use by seniors at a maximum? Estimate the percentage of seniors who had ever used marijuana in that year.

  2. 56.

    1. Use the appropriate line graph to estimate the percentage of seniors who had ever used alcohol in 2015.

    2. Use the appropriate formula to determine the percentage of seniors who had ever used alcohol in 2015. How does this compare with your estimate in part (a)?

    3. Use the appropriate line graph to estimate the percentage of seniors who had ever used marijuana in 2015.

    4. Use the appropriate formula to determine the percentage of seniors who had ever used marijuana in 2015. How does this compare with your estimate in part (c)?

    5. For the period from 1990 through 2019, in which year was alcohol use by seniors at a maximum? What percentage of seniors had ever used alcohol in that year?

Contrary to popular belief, older people do not need less sleep than younger adults. However, the line graphs show that they awaken more often during the night. The numerous awakenings are one reason why some elderly individuals report that sleep is less restful than it had been in the past. Use the line graphs to solve Exercises 5760.

A line graph depicts the average number of awakenings during the night by age and gender.

Source: Stephen Davis and Joseph Palladino, Psychology, 5th Edition, Prentice Hall, 2007

  1. 57. At which age, estimated to the nearest year, do women have the least number of awakenings during the night? What is the average number of awakenings at that age?

  2. 58. At which age do men have the greatest number of awakenings during the night? What is the average number of awakenings at that age?

  3. 59. Estimate, to the nearest tenth, the difference between the average number of awakenings during the night for 25-year-old men and 25-year-old women.

  4. 60. Estimate, to the nearest tenth, the difference between the average number of awakenings during the night for 18-year-old men and 18-year-old women.

Explaining the Concepts

  1. 61. What is the rectangular coordinate system?

  2. 62. Explain how to plot a point in the rectangular coordinate system. Give an example with your explanation.

  3. 63. Explain why (5, 2) and (2, 5) do not represent the same point.

  4. 64. Explain how to graph an equation in the rectangular coordinate system.

  5. 65. What does a [20, 2, 1] by [4, 5, 0.5] viewing rectangle mean?

Technology Exercise

  1. 66. Use a graphing utility to verify each of your hand-drawn graphs in Exercises 1328. Experiment with the settings for the viewing rectangle to make the graph displayed by the graphing utility resemble your hand-drawn graph as much as possible.

Critical Thinking Exercises

MAKE SENSE? In Exercises 6770, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 67. The rectangular coordinate system provides a geometric picture of what an equation in two variables looks like.

  2. 68. There is something wrong with my graphing utility because it is not displaying numbers along the x- and y-axes.

  3. 69. I used the ordered pairs (2, 2), (0, 0), and (2, 2) to graph a straight line.

  4. 70. I used the ordered pairs

         (time of day, calories that I burned)

    to obtain a graph that is a horizontal line.

In Exercises 7174, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 71. If the product of a point’s coordinates is positive, the point must be in quadrant I.

  2. 72. If a point is on the x-axis, it is neither up nor down, so x=0.

  3. 73. If a point is on the y-axis, its x-coordinate must be 0.

  4. 74. The ordered pair (2, 5) satisfies 3y2x=4.

In Exercises 7578, list the quadrant or quadrants satisfying each condition.

  1. 75. xy>0

  2. 76. yx<0

  3. 77. x3>0 and y3<0

  4. 78. x3<0 and y3>0

In Exercises 7982, match the story with the correct figure. The figures are labeled (a), (b), (c), and (d).

  1. 79. As the blizzard got worse, the snow fell harder and harder.

  2. 80. The snow fell more and more softly.

  3. 81. It snowed hard, but then it stopped. After a short time, the snow started falling softly.

  4. 82. It snowed softly, and then it stopped. After a short time, the snow started falling hard.

    1.  

      The graph shows the amount of snowfall versus time, graph shows a concave up curve that slopes upward from origin to the upper right corner.
    2.  

      The graph shows the amount of snowfall versus the time graph shows a line that slopes upward from the origin becomes parallel to the horizontal axis and then again slopes upwards to the upper right corner.
    3.  

      The graph shows the amount of snowfall versus the time, graph shows a line that slopes upward from the origin, becomes parallel to the horizontal axis, and then slopes to the upper right corner.
    4.  

      The graph shows the amount of snowfall versus the time graph shows a concave down curve that slopes upward from origin to the upper right corner.

In Exercises 8387, select the graph that best illustrates each story.

  1. 83. An airplane flew from Miami to San Francisco. [Graphs (c) and (d) are at the top of the next page.]

    1.  

      The graph shows planes' height versus seconds after takeoff graph shows a line that slopes upward from the origin to the upper right corner.
    2.  

      The graph shows planes height versus seconds after takeoff graph shows a line that slopes upward from the origin becomes parallel to the horizontal axis and then slopes downward to meet the horizontal axis near the right end.

    3.  

      The graph shows plane's height versus seconds after takeoff graph shows a line that slopes downward from a point on the vertical axis near the top and then becomes parallel to the horizontal axis in the middle.
    4.  

      The graph shows planes' height versus seconds after takeoff graph shows a line that first moves parallel to the horizontal axis from a point on the vertical axis and then moves vertically downward to meet the horizontal axis near the right end.
  2. 84. At noon, you begin to breathe in.

    1.  

      The graph shows the volume of air in the lungs versus the time afternoon. The graph shows a curve that moves to the right from a vertical axis, first forming a concave down shape and then forming a concave up shape.
    2.  

      The graph shows the volume of air in lungs versus the time afternoon graph shows a curve that moves to the right from a point on the vertical axis, first forming a concave up shape and then forming a concave down shape.
    3.  

      The graph shows the volume of air in the lungs versus the time afternoon graph shows a line that moves parallel to the horizontal axis to the right from a point on the vertical axis.
    4.  

      The graph shows that the volume of air in the lungs versus the afternoon graph shows a line that slopes upward from a point on the vertical axis to the upper right corner.
  3. 85. Measurements are taken of a person’s height from birth to age 100.

    1.  

      The graph shows height versus age graph shows a concave down curve that slopes upward from a point just above the origin.
    2.  

      The graph shows height versus age graph shows a curve that slopes upward from a point just above the origin, first as a concave down curve and then as a concave up curve.
    3.  

      The graph shows height versus age graph shows a concave down curve that moves to the right from a point just above the origin.
    4.  

      The graph shows height versus age graph shows a line that slopes upward from a point just above the origin. The line then becomes parallel to the horizontal axis.
  4. 86. You begin your bike ride by riding down a hill. Then you ride up another hill. Finally, you ride along a level surface before coming to a stop. [Graphs (c) and (d) are at the top of the next column.]

    1.  

      The graph shows speed versus time graph shows a curve that slopes downward from a point on the vertical axis near the top.
    2.  

      The graph shows speed versus time graph shows a curve that slopes upward from the origin as a concave up curve becomes parallel to the horizontal axis and then slopes down to meet the horizontal axis.
    3.  

      The graph shows speed versus time graph plots a curve that slopes upward from the origin as a concave down curve and then slopes downward to meet the horizontal axis near the right end.
    4.  

      The graph shows speed versus time graph plots a curve that slopes downward from a point on the vertical axis near the top as a concave up curve and then becomes parallel to the horizontal axis near the right end.
  5. 87. In Example 6, we used the formula d=1.3n+6 to model the percentage of marriages ending in divorce, d, after n years of marriage for an educational attainment of bachelor’s degree or higher. We can also model the data with the formula

    d=8n6.

    Using a calculator, evaluate each formula for n=5, 10, and 15. Round to the nearest tenth, where necessary. Which model appears to give the better estimates of the percentages shown in Figure 1.11?

Group Exercise

  1. 88. The group should identify three free online graphing calculators. Each group member should graph five equations from this exercise set using each of the three graphing calculators. Then, as a group, discuss what you like or don’t like about the calculators. Based on this discussion, make a list of criteria that you would use in choosing a graphing calculator. Which, if any, of the three graphing calculators would you recommend to fellow students?

Preview Exercises

Exercises 8991 will help you prepare for the material covered in the next section.

  1. 89. Here are two sets of ordered pairs:

    set 1: {(1,5),(2,5)}set 2: {(5,1),(5,2)}.

    In which set is each x-coordinate paired with only one y-coordinate?

  2. 90. Graph y=2x and y=2x+4 in the same rectangular coordinate system. Select integers for x, starting with 2 and ending with 2.

  3. 91. Use the following graph to solve this exercise.

    A graph plots two lines as a V shaped curved intersecting at (0, 1).
    1. What is the y-coordinate when the x-coordinate is 2?

    2. What are the x-coordinates when the y-coordinate is 4?

    3. Describe the x-coordinates of all points on the graph.

    4. Describe the y-coordinates of all points on the graph.

1.1: Exercise Set

1.1 Exercise Set

Practice Exercises

In Exercises 112, plot the given point in a rectangular coordinate system.

  1. 1. (1, 4)

  2. 2. (2, 5)

  3. 3. (2, 3)

  4. 4. (1, 4)

  5. 5. (3, 5)

  6. 6. (4, 2)

  7. 7. (4, 1)

  8. 8. (3, 2)

  9. 9. (4, 0)

  10. 10. (0, 3)

  11. 11. (72, 32)

  12. 12. (52, 32)

Graph each equation in Exercises 1328. Let x=3, 2, 1, 0, 1, 2, and 3.

  1. 13. y=x22

  2. 14. y=x2+2

  3. 15. y=x2

  4. 16. y=x+2

  5. 17. y=2x+1

  6. 18. y=2x4

  7. 19. y=12 x

  8. 20. y=12 x+2

  9. 21. y=2|x|

  10. 22. y=2|x|

  11. 23. y=|x|+1

  12. 24. y=|x|1

  13. 25. y=9x2

  14. 26. y=x2

  15. 27. y=x3

  16. 28. y=x31

In Exercises 2932, match the viewing rectangle with the correct figure. Then label the tick marks in the figure to illustrate this viewing rectangle.

  1. 29. [5, 5, 1] by [5, 5, 1]

  2. 30. [10, 10, 2] by [4, 4, 2]

  3. 31. [20, 80, 10] by [30, 70, 10]

  4. 32. [40, 40, 20] by [1000, 1000, 100]

    1.  

      A graphing calculator screen shows the reviewing rectangle as the x axis with 5 range marks and the y axis with 21 marks. The two axes intersect at the middle mark of each axis.
    2.  

      A graphing calculator screen shows the reviewing rectangle as the x axis with 11 range marks and the y axis with range marks. The two axes intersect at the third mark from the left of x axis and the fourth mark from the bottom of the y axis.
    3.  

      A graphing calculator screenshows the reviewing rectangle as the x axis with 11 range marks and the y axis with 11 range marks. The two axes intersect at the middle mark of each axis.
    4.  

      A graphing calculator screenshows the reviewing rectangle as the x axis with 11 range marks and the y axis with 5 range marks. The two axes intersect at the middle mark of each.

The table of values was generated by a graphing utility with a TABLE feature. Use the table to solve Exercises 3340.

The image depicts the graphing calculator screentable.
  1. 33. Which equation corresponds to Y2 in the table?

    1. y2=x+8

    2. y2=x2

    3. y2=2x

    4. y2=12x

  2. 34. Which equation corresponds to Y1 in the table?

    1. y1=3x

    2. y1=x2

    3. y1=x2

    4. y1=2x

  3. 35. Does the graph of Y2 pass through the origin?

  4. 36. Does the graph of Y1 pass through the origin?

  5. 37. At which point does the graph of Y2 cross the x-axis?

  6. 38. At which point does the graph of Y2 cross the y-axis?

  7. 39. At which points do the graphs of Y1 and Y2 intersect?

  8. 40. For which values of x is Y1=Y2?

In Exercises 4146, use the graph to a. determine the x-intercepts, if any; b. determine the y-intercepts, if any. For each graph, tick marks along the axes represent one unit each.

  1. 41.

    A graph plots a line that rises from the third quadrant to the first quadrant passing through the fourth quadrant. The line intersects the y axis at the fourth range mark below the origin and x axis at the second range mark right to the origin.
  2. 42.

    A graph plots a line that slopes upward from the fourth quadrant to the second quadrant passing through the first quadrant. The line intersects x axis at the first tick mark right to the origin and the y axis at the second tick mark above the origin.

  3. 43.

    A rectangular coordinate system plots a graph of an N shaped curve.
  4. 44.

    A rectangular coordinate system plots a graph of a W shaped curve.
  5. 45.

    A graph plots a left opening parabola with indefinite ends in second and third quadrants, vertex at (1, 0).
  6. 46.

    A graph plots an upward opening parabola with indefinite ends, vertex (0, 2).

Practice PLUS

In Exercises 4750, write each English sentence as an equation in two variables. Then graph the equation.

  1. 47. The y-value is four more than twice the x-value.

  2. 48. The y-value is the difference between four and twice the x-value.

  3. 49. The y-value is three decreased by the square of the x-value.

  4. 50. The y-value is two more than the square of the x-value.

In Exercises 5154, graph each equation.

  1. 51. y=5 (Let x=3, 2, 1, 0, 1, 2, and 3.)

  2. 52. y=1 (Let x=3, 2, 1, 0, 1, 2, and 3.)

  3. 53. y=1x (Let x=2, 1, 12, 13, 13, 12, 1, and 2.)

  4. 54. y= 1x (Let x=2, 1, 12, 13, 13, 12, 1, and 2.)

Application Exercises

The graphs show the percentage of high school seniors who had ever used alcohol or marijuana.

A line graph depicts the alcohol and marijuana use by United States high school seniors.

Source: University of Michigan Institute for Social Research

The data can be described by the following mathematical models:

Percentage of seniors using alcohol, A equals negative 0.9 n + 88. Percentage of seniors using marijuana, M equals 0.1 lower n + 43. n is the number of years after 19 90. A dialogue box pointing both the equation reads, number of year 19 90.

Use this information to solve Exercises 5556.

  1. 55.

    1. Use the appropriate line graph to estimate the percentage of seniors who had ever used marijuana in 2010.

    2. Use the appropriate formula to determine the percentage of seniors who had ever used marijuana in 2010. How does this compare with your estimate in part (a)?

    3. Use the appropriate line graph to estimate the percentage of seniors who had ever used alcohol in 2010.

    4. Use the appropriate formula to determine the percentage of seniors who had ever used alcohol in 2010. How does this compare with your estimate in part (c)?

    5. For the period from 1990 through 2019, in which year was marijuana use by seniors at a maximum? Estimate the percentage of seniors who had ever used marijuana in that year.

  2. 56.

    1. Use the appropriate line graph to estimate the percentage of seniors who had ever used alcohol in 2015.

    2. Use the appropriate formula to determine the percentage of seniors who had ever used alcohol in 2015. How does this compare with your estimate in part (a)?

    3. Use the appropriate line graph to estimate the percentage of seniors who had ever used marijuana in 2015.

    4. Use the appropriate formula to determine the percentage of seniors who had ever used marijuana in 2015. How does this compare with your estimate in part (c)?

    5. For the period from 1990 through 2019, in which year was alcohol use by seniors at a maximum? What percentage of seniors had ever used alcohol in that year?

Contrary to popular belief, older people do not need less sleep than younger adults. However, the line graphs show that they awaken more often during the night. The numerous awakenings are one reason why some elderly individuals report that sleep is less restful than it had been in the past. Use the line graphs to solve Exercises 5760.

A line graph depicts the average number of awakenings during the night by age and gender.

Source: Stephen Davis and Joseph Palladino, Psychology, 5th Edition, Prentice Hall, 2007

  1. 57. At which age, estimated to the nearest year, do women have the least number of awakenings during the night? What is the average number of awakenings at that age?

  2. 58. At which age do men have the greatest number of awakenings during the night? What is the average number of awakenings at that age?

  3. 59. Estimate, to the nearest tenth, the difference between the average number of awakenings during the night for 25-year-old men and 25-year-old women.

  4. 60. Estimate, to the nearest tenth, the difference between the average number of awakenings during the night for 18-year-old men and 18-year-old women.

Explaining the Concepts

  1. 61. What is the rectangular coordinate system?

  2. 62. Explain how to plot a point in the rectangular coordinate system. Give an example with your explanation.

  3. 63. Explain why (5, 2) and (2, 5) do not represent the same point.

  4. 64. Explain how to graph an equation in the rectangular coordinate system.

  5. 65. What does a [20, 2, 1] by [4, 5, 0.5] viewing rectangle mean?

Technology Exercise

  1. 66. Use a graphing utility to verify each of your hand-drawn graphs in Exercises 1328. Experiment with the settings for the viewing rectangle to make the graph displayed by the graphing utility resemble your hand-drawn graph as much as possible.

Critical Thinking Exercises

MAKE SENSE? In Exercises 6770, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 67. The rectangular coordinate system provides a geometric picture of what an equation in two variables looks like.

  2. 68. There is something wrong with my graphing utility because it is not displaying numbers along the x- and y-axes.

  3. 69. I used the ordered pairs (2, 2), (0, 0), and (2, 2) to graph a straight line.

  4. 70. I used the ordered pairs

         (time of day, calories that I burned)

    to obtain a graph that is a horizontal line.

In Exercises 7174, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 71. If the product of a point’s coordinates is positive, the point must be in quadrant I.

  2. 72. If a point is on the x-axis, it is neither up nor down, so x=0.

  3. 73. If a point is on the y-axis, its x-coordinate must be 0.

  4. 74. The ordered pair (2, 5) satisfies 3y2x=4.

In Exercises 7578, list the quadrant or quadrants satisfying each condition.

  1. 75. xy>0

  2. 76. yx<0

  3. 77. x3>0 and y3<0

  4. 78. x3<0 and y3>0

In Exercises 7982, match the story with the correct figure. The figures are labeled (a), (b), (c), and (d).

  1. 79. As the blizzard got worse, the snow fell harder and harder.

  2. 80. The snow fell more and more softly.

  3. 81. It snowed hard, but then it stopped. After a short time, the snow started falling softly.

  4. 82. It snowed softly, and then it stopped. After a short time, the snow started falling hard.

    1.  

      The graph shows the amount of snowfall versus time, graph shows a concave up curve that slopes upward from origin to the upper right corner.
    2.  

      The graph shows the amount of snowfall versus the time graph shows a line that slopes upward from the origin becomes parallel to the horizontal axis and then again slopes upwards to the upper right corner.
    3.  

      The graph shows the amount of snowfall versus the time, graph shows a line that slopes upward from the origin, becomes parallel to the horizontal axis, and then slopes to the upper right corner.
    4.  

      The graph shows the amount of snowfall versus the time graph shows a concave down curve that slopes upward from origin to the upper right corner.

In Exercises 8387, select the graph that best illustrates each story.

  1. 83. An airplane flew from Miami to San Francisco. [Graphs (c) and (d) are at the top of the next page.]

    1.  

      The graph shows planes' height versus seconds after takeoff graph shows a line that slopes upward from the origin to the upper right corner.
    2.  

      The graph shows planes height versus seconds after takeoff graph shows a line that slopes upward from the origin becomes parallel to the horizontal axis and then slopes downward to meet the horizontal axis near the right end.

    3.  

      The graph shows plane's height versus seconds after takeoff graph shows a line that slopes downward from a point on the vertical axis near the top and then becomes parallel to the horizontal axis in the middle.
    4.  

      The graph shows planes' height versus seconds after takeoff graph shows a line that first moves parallel to the horizontal axis from a point on the vertical axis and then moves vertically downward to meet the horizontal axis near the right end.
  2. 84. At noon, you begin to breathe in.

    1.  

      The graph shows the volume of air in the lungs versus the time afternoon. The graph shows a curve that moves to the right from a vertical axis, first forming a concave down shape and then forming a concave up shape.
    2.  

      The graph shows the volume of air in lungs versus the time afternoon graph shows a curve that moves to the right from a point on the vertical axis, first forming a concave up shape and then forming a concave down shape.
    3.  

      The graph shows the volume of air in the lungs versus the time afternoon graph shows a line that moves parallel to the horizontal axis to the right from a point on the vertical axis.
    4.  

      The graph shows that the volume of air in the lungs versus the afternoon graph shows a line that slopes upward from a point on the vertical axis to the upper right corner.
  3. 85. Measurements are taken of a person’s height from birth to age 100.

    1.  

      The graph shows height versus age graph shows a concave down curve that slopes upward from a point just above the origin.
    2.  

      The graph shows height versus age graph shows a curve that slopes upward from a point just above the origin, first as a concave down curve and then as a concave up curve.
    3.  

      The graph shows height versus age graph shows a concave down curve that moves to the right from a point just above the origin.
    4.  

      The graph shows height versus age graph shows a line that slopes upward from a point just above the origin. The line then becomes parallel to the horizontal axis.
  4. 86. You begin your bike ride by riding down a hill. Then you ride up another hill. Finally, you ride along a level surface before coming to a stop. [Graphs (c) and (d) are at the top of the next column.]

    1.  

      The graph shows speed versus time graph shows a curve that slopes downward from a point on the vertical axis near the top.
    2.  

      The graph shows speed versus time graph shows a curve that slopes upward from the origin as a concave up curve becomes parallel to the horizontal axis and then slopes down to meet the horizontal axis.
    3.  

      The graph shows speed versus time graph plots a curve that slopes upward from the origin as a concave down curve and then slopes downward to meet the horizontal axis near the right end.
    4.  

      The graph shows speed versus time graph plots a curve that slopes downward from a point on the vertical axis near the top as a concave up curve and then becomes parallel to the horizontal axis near the right end.
  5. 87. In Example 6, we used the formula d=1.3n+6 to model the percentage of marriages ending in divorce, d, after n years of marriage for an educational attainment of bachelor’s degree or higher. We can also model the data with the formula

    d=8n6.

    Using a calculator, evaluate each formula for n=5, 10, and 15. Round to the nearest tenth, where necessary. Which model appears to give the better estimates of the percentages shown in Figure 1.11?

Group Exercise

  1. 88. The group should identify three free online graphing calculators. Each group member should graph five equations from this exercise set using each of the three graphing calculators. Then, as a group, discuss what you like or don’t like about the calculators. Based on this discussion, make a list of criteria that you would use in choosing a graphing calculator. Which, if any, of the three graphing calculators would you recommend to fellow students?

Preview Exercises

Exercises 8991 will help you prepare for the material covered in the next section.

  1. 89. Here are two sets of ordered pairs:

    set 1: {(1,5),(2,5)}set 2: {(5,1),(5,2)}.

    In which set is each x-coordinate paired with only one y-coordinate?

  2. 90. Graph y=2x and y=2x+4 in the same rectangular coordinate system. Select integers for x, starting with 2 and ending with 2.

  3. 91. Use the following graph to solve this exercise.

    A graph plots two lines as a V shaped curved intersecting at (0, 1).
    1. What is the y-coordinate when the x-coordinate is 2?

    2. What are the x-coordinates when the y-coordinate is 4?

    3. Describe the x-coordinates of all points on the graph.

    4. Describe the y-coordinates of all points on the graph.

1.1: Exercise Set

1.1 Exercise Set

Practice Exercises

In Exercises 112, plot the given point in a rectangular coordinate system.

  1. 1. (1, 4)

  2. 2. (2, 5)

  3. 3. (2, 3)

  4. 4. (1, 4)

  5. 5. (3, 5)

  6. 6. (4, 2)

  7. 7. (4, 1)

  8. 8. (3, 2)

  9. 9. (4, 0)

  10. 10. (0, 3)

  11. 11. (72, 32)

  12. 12. (52, 32)

Graph each equation in Exercises 1328. Let x=3, 2, 1, 0, 1, 2, and 3.

  1. 13. y=x22

  2. 14. y=x2+2

  3. 15. y=x2

  4. 16. y=x+2

  5. 17. y=2x+1

  6. 18. y=2x4

  7. 19. y=12 x

  8. 20. y=12 x+2

  9. 21. y=2|x|

  10. 22. y=2|x|

  11. 23. y=|x|+1

  12. 24. y=|x|1

  13. 25. y=9x2

  14. 26. y=x2

  15. 27. y=x3

  16. 28. y=x31

In Exercises 2932, match the viewing rectangle with the correct figure. Then label the tick marks in the figure to illustrate this viewing rectangle.

  1. 29. [5, 5, 1] by [5, 5, 1]

  2. 30. [10, 10, 2] by [4, 4, 2]

  3. 31. [20, 80, 10] by [30, 70, 10]

  4. 32. [40, 40, 20] by [1000, 1000, 100]

    1.  

      A graphing calculator screen shows the reviewing rectangle as the x axis with 5 range marks and the y axis with 21 marks. The two axes intersect at the middle mark of each axis.
    2.  

      A graphing calculator screen shows the reviewing rectangle as the x axis with 11 range marks and the y axis with range marks. The two axes intersect at the third mark from the left of x axis and the fourth mark from the bottom of the y axis.
    3.  

      A graphing calculator screenshows the reviewing rectangle as the x axis with 11 range marks and the y axis with 11 range marks. The two axes intersect at the middle mark of each axis.
    4.  

      A graphing calculator screenshows the reviewing rectangle as the x axis with 11 range marks and the y axis with 5 range marks. The two axes intersect at the middle mark of each.

The table of values was generated by a graphing utility with a TABLE feature. Use the table to solve Exercises 3340.

The image depicts the graphing calculator screentable.
  1. 33. Which equation corresponds to Y2 in the table?

    1. y2=x+8

    2. y2=x2

    3. y2=2x

    4. y2=12x

  2. 34. Which equation corresponds to Y1 in the table?

    1. y1=3x

    2. y1=x2

    3. y1=x2

    4. y1=2x

  3. 35. Does the graph of Y2 pass through the origin?

  4. 36. Does the graph of Y1 pass through the origin?

  5. 37. At which point does the graph of Y2 cross the x-axis?

  6. 38. At which point does the graph of Y2 cross the y-axis?

  7. 39. At which points do the graphs of Y1 and Y2 intersect?

  8. 40. For which values of x is Y1=Y2?

In Exercises 4146, use the graph to a. determine the x-intercepts, if any; b. determine the y-intercepts, if any. For each graph, tick marks along the axes represent one unit each.

  1. 41.

    A graph plots a line that rises from the third quadrant to the first quadrant passing through the fourth quadrant. The line intersects the y axis at the fourth range mark below the origin and x axis at the second range mark right to the origin.
  2. 42.

    A graph plots a line that slopes upward from the fourth quadrant to the second quadrant passing through the first quadrant. The line intersects x axis at the first tick mark right to the origin and the y axis at the second tick mark above the origin.

  3. 43.

    A rectangular coordinate system plots a graph of an N shaped curve.
  4. 44.

    A rectangular coordinate system plots a graph of a W shaped curve.
  5. 45.

    A graph plots a left opening parabola with indefinite ends in second and third quadrants, vertex at (1, 0).
  6. 46.

    A graph plots an upward opening parabola with indefinite ends, vertex (0, 2).

Practice PLUS

In Exercises 4750, write each English sentence as an equation in two variables. Then graph the equation.

  1. 47. The y-value is four more than twice the x-value.

  2. 48. The y-value is the difference between four and twice the x-value.

  3. 49. The y-value is three decreased by the square of the x-value.

  4. 50. The y-value is two more than the square of the x-value.

In Exercises 5154, graph each equation.

  1. 51. y=5 (Let x=3, 2, 1, 0, 1, 2, and 3.)

  2. 52. y=1 (Let x=3, 2, 1, 0, 1, 2, and 3.)

  3. 53. y=1x (Let x=2, 1, 12, 13, 13, 12, 1, and 2.)

  4. 54. y= 1x (Let x=2, 1, 12, 13, 13, 12, 1, and 2.)

Application Exercises

The graphs show the percentage of high school seniors who had ever used alcohol or marijuana.

A line graph depicts the alcohol and marijuana use by United States high school seniors.

Source: University of Michigan Institute for Social Research

The data can be described by the following mathematical models:

Percentage of seniors using alcohol, A equals negative 0.9 n + 88. Percentage of seniors using marijuana, M equals 0.1 lower n + 43. n is the number of years after 19 90. A dialogue box pointing both the equation reads, number of year 19 90.

Use this information to solve Exercises 5556.

  1. 55.

    1. Use the appropriate line graph to estimate the percentage of seniors who had ever used marijuana in 2010.

    2. Use the appropriate formula to determine the percentage of seniors who had ever used marijuana in 2010. How does this compare with your estimate in part (a)?

    3. Use the appropriate line graph to estimate the percentage of seniors who had ever used alcohol in 2010.

    4. Use the appropriate formula to determine the percentage of seniors who had ever used alcohol in 2010. How does this compare with your estimate in part (c)?

    5. For the period from 1990 through 2019, in which year was marijuana use by seniors at a maximum? Estimate the percentage of seniors who had ever used marijuana in that year.

  2. 56.

    1. Use the appropriate line graph to estimate the percentage of seniors who had ever used alcohol in 2015.

    2. Use the appropriate formula to determine the percentage of seniors who had ever used alcohol in 2015. How does this compare with your estimate in part (a)?

    3. Use the appropriate line graph to estimate the percentage of seniors who had ever used marijuana in 2015.

    4. Use the appropriate formula to determine the percentage of seniors who had ever used marijuana in 2015. How does this compare with your estimate in part (c)?

    5. For the period from 1990 through 2019, in which year was alcohol use by seniors at a maximum? What percentage of seniors had ever used alcohol in that year?

Contrary to popular belief, older people do not need less sleep than younger adults. However, the line graphs show that they awaken more often during the night. The numerous awakenings are one reason why some elderly individuals report that sleep is less restful than it had been in the past. Use the line graphs to solve Exercises 5760.

A line graph depicts the average number of awakenings during the night by age and gender.

Source: Stephen Davis and Joseph Palladino, Psychology, 5th Edition, Prentice Hall, 2007

  1. 57. At which age, estimated to the nearest year, do women have the least number of awakenings during the night? What is the average number of awakenings at that age?

  2. 58. At which age do men have the greatest number of awakenings during the night? What is the average number of awakenings at that age?

  3. 59. Estimate, to the nearest tenth, the difference between the average number of awakenings during the night for 25-year-old men and 25-year-old women.

  4. 60. Estimate, to the nearest tenth, the difference between the average number of awakenings during the night for 18-year-old men and 18-year-old women.

Explaining the Concepts

  1. 61. What is the rectangular coordinate system?

  2. 62. Explain how to plot a point in the rectangular coordinate system. Give an example with your explanation.

  3. 63. Explain why (5, 2) and (2, 5) do not represent the same point.

  4. 64. Explain how to graph an equation in the rectangular coordinate system.

  5. 65. What does a [20, 2, 1] by [4, 5, 0.5] viewing rectangle mean?

Technology Exercise

  1. 66. Use a graphing utility to verify each of your hand-drawn graphs in Exercises 1328. Experiment with the settings for the viewing rectangle to make the graph displayed by the graphing utility resemble your hand-drawn graph as much as possible.

Critical Thinking Exercises

MAKE SENSE? In Exercises 6770, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 67. The rectangular coordinate system provides a geometric picture of what an equation in two variables looks like.

  2. 68. There is something wrong with my graphing utility because it is not displaying numbers along the x- and y-axes.

  3. 69. I used the ordered pairs (2, 2), (0, 0), and (2, 2) to graph a straight line.

  4. 70. I used the ordered pairs

         (time of day, calories that I burned)

    to obtain a graph that is a horizontal line.

In Exercises 7174, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 71. If the product of a point’s coordinates is positive, the point must be in quadrant I.

  2. 72. If a point is on the x-axis, it is neither up nor down, so x=0.

  3. 73. If a point is on the y-axis, its x-coordinate must be 0.

  4. 74. The ordered pair (2, 5) satisfies 3y2x=4.

In Exercises 7578, list the quadrant or quadrants satisfying each condition.

  1. 75. xy>0

  2. 76. yx<0

  3. 77. x3>0 and y3<0

  4. 78. x3<0 and y3>0

In Exercises 7982, match the story with the correct figure. The figures are labeled (a), (b), (c), and (d).

  1. 79. As the blizzard got worse, the snow fell harder and harder.

  2. 80. The snow fell more and more softly.

  3. 81. It snowed hard, but then it stopped. After a short time, the snow started falling softly.

  4. 82. It snowed softly, and then it stopped. After a short time, the snow started falling hard.

    1.  

      The graph shows the amount of snowfall versus time, graph shows a concave up curve that slopes upward from origin to the upper right corner.
    2.  

      The graph shows the amount of snowfall versus the time graph shows a line that slopes upward from the origin becomes parallel to the horizontal axis and then again slopes upwards to the upper right corner.
    3.  

      The graph shows the amount of snowfall versus the time, graph shows a line that slopes upward from the origin, becomes parallel to the horizontal axis, and then slopes to the upper right corner.
    4.  

      The graph shows the amount of snowfall versus the time graph shows a concave down curve that slopes upward from origin to the upper right corner.

In Exercises 8387, select the graph that best illustrates each story.

  1. 83. An airplane flew from Miami to San Francisco. [Graphs (c) and (d) are at the top of the next page.]

    1.  

      The graph shows planes' height versus seconds after takeoff graph shows a line that slopes upward from the origin to the upper right corner.
    2.  

      The graph shows planes height versus seconds after takeoff graph shows a line that slopes upward from the origin becomes parallel to the horizontal axis and then slopes downward to meet the horizontal axis near the right end.

    3.  

      The graph shows plane's height versus seconds after takeoff graph shows a line that slopes downward from a point on the vertical axis near the top and then becomes parallel to the horizontal axis in the middle.
    4.  

      The graph shows planes' height versus seconds after takeoff graph shows a line that first moves parallel to the horizontal axis from a point on the vertical axis and then moves vertically downward to meet the horizontal axis near the right end.
  2. 84. At noon, you begin to breathe in.

    1.  

      The graph shows the volume of air in the lungs versus the time afternoon. The graph shows a curve that moves to the right from a vertical axis, first forming a concave down shape and then forming a concave up shape.
    2.  

      The graph shows the volume of air in lungs versus the time afternoon graph shows a curve that moves to the right from a point on the vertical axis, first forming a concave up shape and then forming a concave down shape.
    3.  

      The graph shows the volume of air in the lungs versus the time afternoon graph shows a line that moves parallel to the horizontal axis to the right from a point on the vertical axis.
    4.  

      The graph shows that the volume of air in the lungs versus the afternoon graph shows a line that slopes upward from a point on the vertical axis to the upper right corner.
  3. 85. Measurements are taken of a person’s height from birth to age 100.

    1.  

      The graph shows height versus age graph shows a concave down curve that slopes upward from a point just above the origin.
    2.  

      The graph shows height versus age graph shows a curve that slopes upward from a point just above the origin, first as a concave down curve and then as a concave up curve.
    3.  

      The graph shows height versus age graph shows a concave down curve that moves to the right from a point just above the origin.
    4.  

      The graph shows height versus age graph shows a line that slopes upward from a point just above the origin. The line then becomes parallel to the horizontal axis.
  4. 86. You begin your bike ride by riding down a hill. Then you ride up another hill. Finally, you ride along a level surface before coming to a stop. [Graphs (c) and (d) are at the top of the next column.]

    1.  

      The graph shows speed versus time graph shows a curve that slopes downward from a point on the vertical axis near the top.
    2.  

      The graph shows speed versus time graph shows a curve that slopes upward from the origin as a concave up curve becomes parallel to the horizontal axis and then slopes down to meet the horizontal axis.
    3.  

      The graph shows speed versus time graph plots a curve that slopes upward from the origin as a concave down curve and then slopes downward to meet the horizontal axis near the right end.
    4.  

      The graph shows speed versus time graph plots a curve that slopes downward from a point on the vertical axis near the top as a concave up curve and then becomes parallel to the horizontal axis near the right end.
  5. 87. In Example 6, we used the formula d=1.3n+6 to model the percentage of marriages ending in divorce, d, after n years of marriage for an educational attainment of bachelor’s degree or higher. We can also model the data with the formula

    d=8n6.

    Using a calculator, evaluate each formula for n=5, 10, and 15. Round to the nearest tenth, where necessary. Which model appears to give the better estimates of the percentages shown in Figure 1.11?

Group Exercise

  1. 88. The group should identify three free online graphing calculators. Each group member should graph five equations from this exercise set using each of the three graphing calculators. Then, as a group, discuss what you like or don’t like about the calculators. Based on this discussion, make a list of criteria that you would use in choosing a graphing calculator. Which, if any, of the three graphing calculators would you recommend to fellow students?

Preview Exercises

Exercises 8991 will help you prepare for the material covered in the next section.

  1. 89. Here are two sets of ordered pairs:

    set 1: {(1,5),(2,5)}set 2: {(5,1),(5,2)}.

    In which set is each x-coordinate paired with only one y-coordinate?

  2. 90. Graph y=2x and y=2x+4 in the same rectangular coordinate system. Select integers for x, starting with 2 and ending with 2.

  3. 91. Use the following graph to solve this exercise.

    A graph plots two lines as a V shaped curved intersecting at (0, 1).
    1. What is the y-coordinate when the x-coordinate is 2?

    2. What are the x-coordinates when the y-coordinate is 4?

    3. Describe the x-coordinates of all points on the graph.

    4. Describe the y-coordinates of all points on the graph.

Section 1.2: Basics of Functions and Their Graphs

Section 1.2 Basics of Functions and Their Graphs

Learning Objectives

What You’ll Learn

  1. 1 Find the domain and range of a relation.

  2. 2 Determine whether a relation is a function.

  3. 3 Determine whether an equation represents a function.

  4. 4 Evaluate a function.

  5. 5 Graph functions by plotting points.

  6. 6 Use the vertical line test to identify functions.

  7. 7 Obtain information about a function from its graph.

  8. 8 Identify the domain and range of a function from its graph.

  9. 9 Identify intercepts from a function’s graph.

The number of T cells in a person with HIV is a function of time after infection. In this section, you will be introduced to the basics of functions and their graphs. We will analyze the graph of a function using an example that illustrates the progression of HIV and T cell count. Much of our work in this course will be devoted to the important topic of functions and how they model your world.

Magnified 6000 times, this color-scanned image shows a T-lymphocyte blood cell (green) infected with the HIV virus (red). Depletion of the number of T cells causes destruction of the immune system.

Objective 1: Find the domain and range of a relation

Relations

  1. Objective 1 Find the domain and range of a relation.

Watch Video

Forbes magazine published a list of the highest-paid TV actors and actresses in 2018. The results are shown in Figure 1.13.

Figure 1.13

A vertical bar graph titled, highest paid T V actors and actresses in 20 18.

Source: Forbes

Figure 1.13 Full Alternative Text

The graph indicates a correspondence between a TV actor or actress and that person’s earnings, in millions of dollars. We can write this correspondence using a set of ordered pairs:

An equation.
1.2-52 Full Alternative Text

The mathematical term for a set of ordered pairs is a relation.

Definition of a Relation

A relation is any set of ordered pairs. The set of all first components of the ordered pairs is called the domain of the relation and the set of all second components is called the range of the relation.

Example 1 Finding the Domain and Range of a Relation

Find the domain and range of the relation:

{(Vergara, 43), (Parsons, 28), (Galecki, 27), (Cuoco, 26), (Helberg, 26)}.

Solution

The domain is the set of all first components. Thus, the domain is

{Vergara, Parsons, Galecki, Cuoco, Helberg}.

The range is the set of all second components. Thus, the range is

Left brace 43 comma 28 comma 27 comma 26 right brace. Although Cuoco and Helberg both earned $26 million, it is not necessary to list 26 twice.

Check Point 1

  • Find the domain and range of the relation:

    {(0, 187), (1, 212), (2, 225), (3, 238)}.

As you worked Check Point 1, did you wonder if there was a rule that assigned the “inputs” in the domain to the “outputs” in the range? For example, for the ordered pair (2, 225), how does the output 225 depend on the input 2? The ordered pair is based on the data in Figure 1.14(a), which shows the number of smartphone users, in millions, in the United States.

Figure 1.14(a)

A bar graph.

Figure 1.14(b)

A dot plot.

Source: New York Times Upfront

In Figure 1.14(b), we used the data for the number of smartphone users to create the following ordered pairs:

(years after 2015, the number of smartphoneusers, in millions).

Consider, for example, the ordered pair (2, 225).

Left parenthesis 2, 225 right parenthesis. 2 represents 2 years after 20 15, or in 20 17, and 225 represents that there were 225 million smartphone users in the United States.

The four points in Figure 1.14(b) visually represent the relation formed from the data. Another way to visually represent the relation is as follows:

A mapping diagram. The domain values are as follows. 0, 1, 2, and 3. The range values are as follows. 187, 212, 225, and 238. The mappings from the domain to the range are as follows. 0, 187. 1, 212. 2, 225. 3, 238.
Objective 1: Find the domain and range of a relation

Relations

  1. Objective 1 Find the domain and range of a relation.

Watch Video

Forbes magazine published a list of the highest-paid TV actors and actresses in 2018. The results are shown in Figure 1.13.

Figure 1.13

A vertical bar graph titled, highest paid T V actors and actresses in 20 18.

Source: Forbes

Figure 1.13 Full Alternative Text

The graph indicates a correspondence between a TV actor or actress and that person’s earnings, in millions of dollars. We can write this correspondence using a set of ordered pairs:

An equation.
1.2-52 Full Alternative Text

The mathematical term for a set of ordered pairs is a relation.

Definition of a Relation

A relation is any set of ordered pairs. The set of all first components of the ordered pairs is called the domain of the relation and the set of all second components is called the range of the relation.

Example 1 Finding the Domain and Range of a Relation

Find the domain and range of the relation:

{(Vergara, 43), (Parsons, 28), (Galecki, 27), (Cuoco, 26), (Helberg, 26)}.

Solution

The domain is the set of all first components. Thus, the domain is

{Vergara, Parsons, Galecki, Cuoco, Helberg}.

The range is the set of all second components. Thus, the range is

Left brace 43 comma 28 comma 27 comma 26 right brace. Although Cuoco and Helberg both earned $26 million, it is not necessary to list 26 twice.

Check Point 1

  • Find the domain and range of the relation:

    {(0, 187), (1, 212), (2, 225), (3, 238)}.

As you worked Check Point 1, did you wonder if there was a rule that assigned the “inputs” in the domain to the “outputs” in the range? For example, for the ordered pair (2, 225), how does the output 225 depend on the input 2? The ordered pair is based on the data in Figure 1.14(a), which shows the number of smartphone users, in millions, in the United States.

Figure 1.14(a)

A bar graph.

Figure 1.14(b)

A dot plot.

Source: New York Times Upfront

In Figure 1.14(b), we used the data for the number of smartphone users to create the following ordered pairs:

(years after 2015, the number of smartphoneusers, in millions).

Consider, for example, the ordered pair (2, 225).

Left parenthesis 2, 225 right parenthesis. 2 represents 2 years after 20 15, or in 20 17, and 225 represents that there were 225 million smartphone users in the United States.

The four points in Figure 1.14(b) visually represent the relation formed from the data. Another way to visually represent the relation is as follows:

A mapping diagram. The domain values are as follows. 0, 1, 2, and 3. The range values are as follows. 187, 212, 225, and 238. The mappings from the domain to the range are as follows. 0, 187. 1, 212. 2, 225. 3, 238.
Objective 2: Determine whether a relation is a function

Functions

  1. Objective 2 Determine whether a relation is a function.

Watch Video

Table 1.1, based on our earlier discussion, shows the highest-paid TV actors and actresses and their earnings in 2018, in millions of dollars. We’ve used this information to define two relations.

Table 1.1 Highest-Paid TV Actors and Actresses

Actor/Actress Earnings (millions of dollars)
Vergara 43
Parsons 28
Galecki 27
Cuoco 26
Helberg 26

Figure 1.15(a) shows a correspondence between actors and actresses and their earnings. Figure 1.15(b) shows a correspondence between earnings and actors and actresses.

Figure 1.15(a) Actors and actresses correspond to earnings.

A mapping diagram.
Figure 1.15 (a) Full Alternative Text

Figure 1.15(b) Earnings correspond to actors and actresses.

A mapping diagram.
Figure 1.15 (b) Full Alternative Text

A relation in which each member of the domain corresponds to exactly one member of the range is a function. Can you see that the relation in Figure 1.15(a) is a function? Each actor or actress in the domain corresponds to exactly one earnings amount in the range. If we know the actor or actress, we can be sure of his or her earnings. Notice that more than one element in the domain can correspond to the same element in the range: Cuoco and Helberg both earned $26 million.

Is the relation in Figure 1.15(b) a function? Does each member of the domain correspond to precisely one member of the range? This relation is not a function because there is a member of the domain that corresponds to two different members of the range:

(26, Cuoco)(26, Helberg).

The member of the domain 26 corresponds to both Cuoco and Helberg. If we know that the earnings are $26 million, we cannot be sure of the actor or actress. Because a function is a relation in which no two ordered pairs have the same first component and different second components, the ordered pairs (26, Cuoco) and (26, Helberg) are not ordered pairs of a function.

The following ordered pairs have the same first component but different second components. (26, Cuoco) and (26, Helberg).

Definition of a Function

A function is a correspondence from a first set, called the domain, to a second set, called the range, such that each element in the domain corresponds to exactly one element in the range.

Example 2 illustrates that not every correspondence between sets is a function.

Example 2 Determining Whether a Relation Is a Function

Determine whether each relation is a function:

  1. {(1, 6), (2, 6), (3, 8), (4, 9)}

  2. {(6, 1), (6, 2), (8, 3), (9, 4)}.

Solution

We begin by making a figure for each relation that shows the domain and the range (Figure 1.16).

Figure 1.16(a)

A mapping diagram. The domain values are as follows. 1, 2, 3, and 4. The range values are 6, 8, and 9. The mappings from the domain to the range are as follows. 1, 6. 2, 6. 3, 8. 4, 9.

Figure 1.16(b)

A mapping diagram. The domain values are as follows. 6, 8, and 9. The range values are 1, 2, 3, and 4. The mappings from the domain to the range are as follows. 6, 1. 6, 2. 8, 3. 9, 4.
  1. Figure 1.16(a) shows that every element in the domain corresponds to exactly one element in the range. The element 1 in the domain corresponds to the element 6 in the range. Furthermore, 2 corresponds to 6, 3 corresponds to 8, and 4 corresponds to 9. No two ordered pairs in the given relation have the same first component and different second components. Thus, the relation is a function.

  2. Figure 1.16(b) shows that 6 corresponds to both 1 and 2. If any element in the domain corresponds to more than one element in the range, the relation is not a function. This relation is not a function; two ordered pairs have the same first component and different second components.

    The following ordered pairs have the same first component but different second components. (6, 1) and (6, 2).

Look at Figure 1.16(a) again. The fact that 1 and 2 in the domain correspond to the same number, 6, in the range does not violate the definition of a function. A function can have two different first components with the same second component. By contrast, a relation is not a function when two different ordered pairs have the same first component and different second components. Thus, the relation in Figure 1.16(b) is not a function.

Check Point 2

  • Determine whether each relation is a function:

    1. {(1, 2), (3, 4), (5, 6), (5, 8)}

    2. {(1, 2), (3, 4), (6, 5), (8, 5)}.

Objective 3: Determine whether an equation represents a function

Functions as Equations

  1. Objective 3 Determine whether an equation represents a function.

Watch Video

Functions are usually given in terms of equations rather than as sets of ordered pairs. For example, here is an equation that models the percentage of first-year college women claiming no religious affiliation as a function of time:

y=0.012x20.2x+8.7.

The variable x represents the number of years after 1970. The variable y represents the percentage of first-year college women claiming no religious affiliation. The variable y is a function of the variable x. For each value of x, there is one and only one value of y. The variable x is called the independent variable because it can be assigned any value from the domain. Thus, x can be assigned any nonnegative integer representing the number of years after 1970. The variable y is called the dependent variable because its value depends on x. The percentage claiming no religious affiliation depends on the number of years after 1970. The value of the dependent variable, y, is calculated after selecting a value for the independent variable, x.

We have seen that not every set of ordered pairs defines a function. Similarly, not all equations with the variables x and y define functions. If an equation is solved for y and more than one value of y can be obtained for a given x, then the equation does not define y as a function of x.

Example 3 Determining Whether an Equation Represents a Function

Determine whether each equation defines y as a function of x:

  1. x2+y=4

  2. x2+y2=4.

Solution

Solve each equation for y in terms of x. If two or more values of y can be obtained for a given x, the equation is not a function.

  1.  

    x2+y=4This is the given equation.x2+yx2=4x2Solve for y  by subtracting x2 from both sides.y=4x2Simplify.

    From this last equation we can see that for each value of x, there is one and only one value of y. For example, if x=1, then y=412=3. The equation defines y as a function of x.

  2.  

    x2+y2=4This is the given equation.x2+y2x2=4x2Isolate y2 by subtracting x2 from both sides.y2=4x2Simplify.y=±4x2Apply the square root property: If u2=d,then u=±d.

    The ± in this last equation shows that for certain values of x (all values between 2 and 2), there are two values of y. For example, if x=1, then y=±412=±3. For this reason, the equation does not define y as a function of x.

Check Point 3

  • Solve each equation for y and then determine whether the equation defines y as a function of x:

    1. 2x+y=6

    2. x2+y2=1.

Objective 4: Evaluate a function

Function Notation

  1. Objective 4 Evaluate a function.

Watch Video

If an equation in x and y gives one and only one value of y for each value of x, then the variable y is a function of the variable x. When an equation represents a function, the function is often named by a letter such as f, g, h, F, G, or H. Any letter can be used to name a function. Suppose that f names a function. Think of the domain as the set of the function’s inputs and the range as the set of the function’s outputs. As shown in Figure 1.17, input is represented by x and the output by f(x). The special notation f(x), read “f of x” or “f at x,” represents the value of the function at the number x.

Figure 1.17 A “function machine” with inputs and outputs

A function machine f is given an input x and produces an output f of x.

Let’s make this clearer by considering a specific example. The equation y=0.012x20.2x+8.7 models the percentage of college women claiming no religious affiliation, y, x years after 1970. This equation defines y as a function of x. We’ll name the function f. Now, we can apply our new function notation.

The input is x, and the output is f of x. The equation is f of x = 0.012 x squared minus 0.2 x + 8.7. A dialogue box pointing at the equation reads, we read this equation as f of x equals 0.012 x squared minus 0.2 x + 8.7.

Suppose we are interested in finding f(30), the function’s output when the input is 30. To find the value of the function at 30, we substitute 30 for x. We are evaluating the function at 30.

f(x)=0.012x20.2x+8.7This is the given function.f(30)=0.012(30)20.2(30)+8.7Replace each occurrence of x with 30.=0.012(900)0.2(30)+8.7Evaluate the exponential expression:302=3030=900.=10.86+8.7Perform the multiplications.f(30)=13.5Subtract and add from left to right.

The statement f(30)=13.5, read “f of 30 equals 13.5,” tells us that the value of the function at 30 is 13.5. When the function’s input is 30, its output is 13.5. Figure 1.18 illustrates the input and output in terms of a function machine.

f of 30 = 13.5. A dialogue box pointing at 30 reads represents 30 years after 19 70, or in 2000, and A dialogue box pointing at 13.5 reads, 13.5 represents that 13.5% of the first year college women claimed no religious affiliation.

Figure 1.18 A function machine at work

A function machine, f of x = 0.012 x squared minus 0.2 x + 8.7, is given an input x = 30. The process that takes place in the machine is 0.012 times 30 squared minus 0.2 times 30 + 8.7. The output produced is f of 30 = 13.5.

In 2000, 13.2% actually claimed nonaffiliation, so our function that models the data slightly overestimates the percent for 2000.

We used f(x)=0.012x20.2x+8.7 to find f(30). To find other function values, such as f(40) or f(55), substitute the specified input value, 40 or 55, for x in the function’s equation.

If a function is named f and x represents the independent variable, the notation f(x) corresponds to the y-value for a given x. Thus,

f(x)=0.012x20.2x+8.7andy=0.012x20.2x+8.7

define the same function. This function may be written as

y=f(x)=0.012x20.2x+8.7.

Example 4 Evaluating a Function

If f(x)=x2+3x+5, evaluate each of the following:

  1. f(2)

  2. f(x+3)

  3. f(x).

Solution

We substitute 2, x+3, and x for x in the equation for f. When replacing x with a variable or an algebraic expression, you might find it helpful to think of the function’s equation as

f(x)=x2+3x+5.
  1. We find f(2) by substituting 2 for x in the equation.

    f(2)=22+32+5=4+6+5=15

    Thus, f(2)=15.

  2. We find f(x+3) by substituting x+3 for x in the equation.

    f(x+3)=(x+3 )2+3(x+3)+5

    Equivalently,

    f of start expression x + 3 end expression right parenthesis = left parenthesis x + 3 right parenthesis squared + 3 left parenthesis x + 3 right parenthesis + 5. Arrows extends from 3 to x and 3.
  3. We find f(x) by substituting x for x in the equation.

    f(x)=(x)2+3(x)+5

    Equivalently,

    f(x)=(x)2+3(x)+5  f (..)=x23x+5.

Check Point 4

  • If f(x)=x22x+7, evaluate each of the following:

    1. f(5)

    2. f(x+4)

    3. f(x).

Objective 4: Evaluate a function

Function Notation

  1. Objective 4 Evaluate a function.

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If an equation in x and y gives one and only one value of y for each value of x, then the variable y is a function of the variable x. When an equation represents a function, the function is often named by a letter such as f, g, h, F, G, or H. Any letter can be used to name a function. Suppose that f names a function. Think of the domain as the set of the function’s inputs and the range as the set of the function’s outputs. As shown in Figure 1.17, input is represented by x and the output by f(x). The special notation f(x), read “f of x” or “f at x,” represents the value of the function at the number x.

Figure 1.17 A “function machine” with inputs and outputs

A function machine f is given an input x and produces an output f of x.

Let’s make this clearer by considering a specific example. The equation y=0.012x20.2x+8.7 models the percentage of college women claiming no religious affiliation, y, x years after 1970. This equation defines y as a function of x. We’ll name the function f. Now, we can apply our new function notation.

The input is x, and the output is f of x. The equation is f of x = 0.012 x squared minus 0.2 x + 8.7. A dialogue box pointing at the equation reads, we read this equation as f of x equals 0.012 x squared minus 0.2 x + 8.7.

Suppose we are interested in finding f(30), the function’s output when the input is 30. To find the value of the function at 30, we substitute 30 for x. We are evaluating the function at 30.

f(x)=0.012x20.2x+8.7This is the given function.f(30)=0.012(30)20.2(30)+8.7Replace each occurrence of x with 30.=0.012(900)0.2(30)+8.7Evaluate the exponential expression:302=3030=900.=10.86+8.7Perform the multiplications.f(30)=13.5Subtract and add from left to right.

The statement f(30)=13.5, read “f of 30 equals 13.5,” tells us that the value of the function at 30 is 13.5. When the function’s input is 30, its output is 13.5. Figure 1.18 illustrates the input and output in terms of a function machine.

f of 30 = 13.5. A dialogue box pointing at 30 reads represents 30 years after 19 70, or in 2000, and A dialogue box pointing at 13.5 reads, 13.5 represents that 13.5% of the first year college women claimed no religious affiliation.

Figure 1.18 A function machine at work

A function machine, f of x = 0.012 x squared minus 0.2 x + 8.7, is given an input x = 30. The process that takes place in the machine is 0.012 times 30 squared minus 0.2 times 30 + 8.7. The output produced is f of 30 = 13.5.

In 2000, 13.2% actually claimed nonaffiliation, so our function that models the data slightly overestimates the percent for 2000.

We used f(x)=0.012x20.2x+8.7 to find f(30). To find other function values, such as f(40) or f(55), substitute the specified input value, 40 or 55, for x in the function’s equation.

If a function is named f and x represents the independent variable, the notation f(x) corresponds to the y-value for a given x. Thus,

f(x)=0.012x20.2x+8.7andy=0.012x20.2x+8.7

define the same function. This function may be written as

y=f(x)=0.012x20.2x+8.7.

Example 4 Evaluating a Function

If f(x)=x2+3x+5, evaluate each of the following:

  1. f(2)

  2. f(x+3)

  3. f(x).

Solution

We substitute 2, x+3, and x for x in the equation for f. When replacing x with a variable or an algebraic expression, you might find it helpful to think of the function’s equation as

f(x)=x2+3x+5.
  1. We find f(2) by substituting 2 for x in the equation.

    f(2)=22+32+5=4+6+5=15

    Thus, f(2)=15.

  2. We find f(x+3) by substituting x+3 for x in the equation.

    f(x+3)=(x+3 )2+3(x+3)+5

    Equivalently,

    f of start expression x + 3 end expression right parenthesis = left parenthesis x + 3 right parenthesis squared + 3 left parenthesis x + 3 right parenthesis + 5. Arrows extends from 3 to x and 3.
  3. We find f(x) by substituting x for x in the equation.

    f(x)=(x)2+3(x)+5

    Equivalently,

    f(x)=(x)2+3(x)+5  f (..)=x23x+5.

Check Point 4

  • If f(x)=x22x+7, evaluate each of the following:

    1. f(5)

    2. f(x+4)

    3. f(x).

Objective 5: Graph functions by plotting points

Graphs of Functions

  1. Objective 5 Graph functions by plotting points.

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The graph of a function is the graph of its ordered pairs. For example, the graph of f(x)=2x is the set of points (x, y) in the rectangular coordinate system satisfying y=2x. Similarly, the graph of g(x)=2x+4 is the set of points (x, y) in the rectangular coordinate system satisfying the equation y=2x+4. In the next example, we graph both of these functions in the same rectangular coordinate system.

Example 5 Graphing Functions

Graph the functions f(x)=2x and g(x)=2x+4 in the same rectangular coordinate system. Select integers for x, starting with 2 and ending with 2.

Solution

We begin by setting up a partial table of coordinates for each function. Then we plot the five points in each table and connect them, as shown in Figure 1.19 on the next page. The graph of each function is a straight line. Do you see a relationship between the two graphs? The graph of g is the graph of f shifted vertically up by 4 units.

A table lists the values of x, f of x = 2 x, and (x, y) or (x, f of x).
A table lists the values of x, g of x = 2 x + 4, and (x, y) or (x, g of x).

Figure 1.19

A graph plots two lines for f of x = 2 x and g of x = 2 x + 4. f of x is moved 4 units up to obtain g of x.

The graphs in Example 5 are straight lines. All functions with equations of the form f(x)=mx+b graph as straight lines. Such functions, called linear functions, will be discussed in detail in Section 1.4.

Check Point 5

  • Graph the functions f(x)=2x and g(x)=2x3 in the same rectangular coordinate system. Select integers for x, starting with 2 and ending with 2. How is the graph of g related to the graph of f?

The Vertical Line Test

Not every graph in the rectangular coordinate system is the graph of a function. The definition of a function specifies that no value of x can be paired with two or more different values of y. Consequently, if a graph contains two or more different points with the same first coordinate, the graph cannot represent a function. This is illustrated in Figure 1.20. Observe that points sharing a common first coordinate are vertically above or below each other.

Figure 1.20 y is not a function of x because 0 is paired with three values of y, namely, 1, 0, and 1.

A graph plots a curve with indefinite ends.

This observation is the basis of a useful test for determining whether a graph defines y as a function of x. The test is called the vertical line test.

Objective 5: Graph functions by plotting points

Graphs of Functions

  1. Objective 5 Graph functions by plotting points.

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The graph of a function is the graph of its ordered pairs. For example, the graph of f(x)=2x is the set of points (x, y) in the rectangular coordinate system satisfying y=2x. Similarly, the graph of g(x)=2x+4 is the set of points (x, y) in the rectangular coordinate system satisfying the equation y=2x+4. In the next example, we graph both of these functions in the same rectangular coordinate system.

Example 5 Graphing Functions

Graph the functions f(x)=2x and g(x)=2x+4 in the same rectangular coordinate system. Select integers for x, starting with 2 and ending with 2.

Solution

We begin by setting up a partial table of coordinates for each function. Then we plot the five points in each table and connect them, as shown in Figure 1.19 on the next page. The graph of each function is a straight line. Do you see a relationship between the two graphs? The graph of g is the graph of f shifted vertically up by 4 units.

A table lists the values of x, f of x = 2 x, and (x, y) or (x, f of x).
A table lists the values of x, g of x = 2 x + 4, and (x, y) or (x, g of x).

Figure 1.19

A graph plots two lines for f of x = 2 x and g of x = 2 x + 4. f of x is moved 4 units up to obtain g of x.

The graphs in Example 5 are straight lines. All functions with equations of the form f(x)=mx+b graph as straight lines. Such functions, called linear functions, will be discussed in detail in Section 1.4.

Check Point 5

  • Graph the functions f(x)=2x and g(x)=2x3 in the same rectangular coordinate system. Select integers for x, starting with 2 and ending with 2. How is the graph of g related to the graph of f?

The Vertical Line Test

Not every graph in the rectangular coordinate system is the graph of a function. The definition of a function specifies that no value of x can be paired with two or more different values of y. Consequently, if a graph contains two or more different points with the same first coordinate, the graph cannot represent a function. This is illustrated in Figure 1.20. Observe that points sharing a common first coordinate are vertically above or below each other.

Figure 1.20 y is not a function of x because 0 is paired with three values of y, namely, 1, 0, and 1.

A graph plots a curve with indefinite ends.

This observation is the basis of a useful test for determining whether a graph defines y as a function of x. The test is called the vertical line test.

Objective 7: Obtain information about a function from its graph

Obtaining Information from Graphs

  1. Objective 7 Obtain information about a function from its graph.

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You can obtain information about a function from its graph. At the right or left of a graph, you will find closed dots, open dots, or arrows.

Example 7 Analyzing the Graph of a Function

The human immunodeficiency virus, or HIV, infects and kills helper T cells. Because T cells stimulate the immune system to produce antibodies, their destruction disables the body’s defenses against other pathogens. By counting the number of T cells that remain active in the body, the progression of HIV can be monitored. The fewer helper T cells, the more advanced the disease. Without the drugs that are now used to inhibit the progression of the virus, Figure 1.21 shows a graph that is used to monitor the average progression of the disease. The number of T cells, f(x), is a function of time after infection, x.

Figure 1.21

A graph titled T cell count and H I V infection plots the curve of y = f of x.

Source: B. E. Pruitt et al., Human Sexuality, Prentice Hall, 2007

  1. Explain why f represents the graph of a function.

  2. Use the graph to find and interpret f(8).

  3. For what value of x is f(x)=350?

  4. Describe the general trend shown by the graph.

Solution

  1. No vertical line can be drawn that intersects the graph of f more than once. By the vertical line test, f represents the graph of a function.

  2. To find f(8), or f of 8, we locate 8 on the x-axis. Figure 1.22 shows the point on the graph of f for which 8 is the first coordinate. From this point, we look to the y-axis to find the corresponding y-coordinate. We see that the y-coordinate is 200. Thus,

    f(8)=200.

    Figure 1.22 Finding f(8)

    A graph plots the curve of y = f of x.

    When the time after infection is 8 years, the T cell count is 200 cells per milliliter of blood. (AIDS clinical diagnosis is given at a T cell count of 200 or below.)

  3. To find the value of x for which f(x)=350, we find the approximate location of 350 on the y-axis. Figure 1.23 shows that there is one point on the graph of f for which 350 is the second coordinate. From this point, we look to the x-axis to find the corresponding x-coordinate. We see that the x-coordinate is 6. Thus,

    f(x)=350 forx=6.

    Figure 1.23 Finding x for which f(x)=350

    A graph plots the curve of y = f of x.

    A T cell count of 350 occurs 6 years after infection.

  4. Figure 1.24 uses voice balloons to describe the general trend shown by the graph.

    Figure 1.24 Describing changing T cell count over time in a person infected with HIV

    A graph plots the T cell count versus the time after infection as a curve.

Check Point 7

    1. Use the graph of f in Figure 1.21 to find f(5).

    2. For what value of x is f(x)=100?

    3. Estimate the minimum T cell count during the asymptomatic stage.

Objective 7: Obtain information about a function from its graph

Obtaining Information from Graphs

  1. Objective 7 Obtain information about a function from its graph.

Watch Video

You can obtain information about a function from its graph. At the right or left of a graph, you will find closed dots, open dots, or arrows.

Example 7 Analyzing the Graph of a Function

The human immunodeficiency virus, or HIV, infects and kills helper T cells. Because T cells stimulate the immune system to produce antibodies, their destruction disables the body’s defenses against other pathogens. By counting the number of T cells that remain active in the body, the progression of HIV can be monitored. The fewer helper T cells, the more advanced the disease. Without the drugs that are now used to inhibit the progression of the virus, Figure 1.21 shows a graph that is used to monitor the average progression of the disease. The number of T cells, f(x), is a function of time after infection, x.

Figure 1.21

A graph titled T cell count and H I V infection plots the curve of y = f of x.

Source: B. E. Pruitt et al., Human Sexuality, Prentice Hall, 2007

  1. Explain why f represents the graph of a function.

  2. Use the graph to find and interpret f(8).

  3. For what value of x is f(x)=350?

  4. Describe the general trend shown by the graph.

Solution

  1. No vertical line can be drawn that intersects the graph of f more than once. By the vertical line test, f represents the graph of a function.

  2. To find f(8), or f of 8, we locate 8 on the x-axis. Figure 1.22 shows the point on the graph of f for which 8 is the first coordinate. From this point, we look to the y-axis to find the corresponding y-coordinate. We see that the y-coordinate is 200. Thus,

    f(8)=200.

    Figure 1.22 Finding f(8)

    A graph plots the curve of y = f of x.

    When the time after infection is 8 years, the T cell count is 200 cells per milliliter of blood. (AIDS clinical diagnosis is given at a T cell count of 200 or below.)

  3. To find the value of x for which f(x)=350, we find the approximate location of 350 on the y-axis. Figure 1.23 shows that there is one point on the graph of f for which 350 is the second coordinate. From this point, we look to the x-axis to find the corresponding x-coordinate. We see that the x-coordinate is 6. Thus,

    f(x)=350 forx=6.

    Figure 1.23 Finding x for which f(x)=350

    A graph plots the curve of y = f of x.

    A T cell count of 350 occurs 6 years after infection.

  4. Figure 1.24 uses voice balloons to describe the general trend shown by the graph.

    Figure 1.24 Describing changing T cell count over time in a person infected with HIV

    A graph plots the T cell count versus the time after infection as a curve.

Check Point 7

    1. Use the graph of f in Figure 1.21 to find f(5).

    2. For what value of x is f(x)=100?

    3. Estimate the minimum T cell count during the asymptomatic stage.

Objective 8: Identify the domain and range of a function from its graph

Identifying Domain and Range from a Function’s Graph

  1. Objective 8 Identify the domain and range of a function from its graph.

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Figure 1.25 illustrates how the graph of a function is used to determine the function’s domain and its range.

The domain colon set of inputs is labeled, found on the x axis. The range colon set of outputs is labeled, found on the y axis.

Figure 1.25 Domain and range of f

The graph shows a concave down curve.

Figure 1.25 Full Alternative Text

Let’s apply these ideas to the graph of the function shown in Figure 1.26. To find the domain, look for all the inputs on the x-axis that correspond to points on the graph. Can you see that they extend from 4 to 2, inclusive? The function’s domain can be represented as follows:

Examples of using set builder notation and interval notation.
1.2-73 Full Alternative Text

Figure 1.26 Domain and range of f

The graph of y = f of x is a semi circle that rises from (negative 4, 1) to (negative 1, 3) and then falls to (2, 1). The two ends of the curve are marked by closed points.

Figure 1.26 Full Alternative Text

To find the range, look for all the outputs on the y-axis that correspond to points on the graph. They extend from 1 to 4, inclusive. The function’s range can be represented as follows:

Examples of using set builder notation and interval notation.
1.2-74 Full Alternative Text

Example 8 Identifying the Domain and Range of a Function from Its Graph

Use the graph of each function to identify its domain and its range.

  1.  

    The graph of y = f of x is a line that rises from the closed point at (negative 2, 0) to (0, 1) and then to the closed point at (1, 3).
  2.  

    The graph of y = f of x is a line that rises from the open point at (negative 3, 1) to the closed point at (2, 2).
  3.  

    The graph of y = f of x is a curve that falls from the closed point at (negative 2, 5) to (0, 1), and then rises to the open point at (1, 2).
  4.  

    The graph of y = f of x is a curve that rises from the closed point at (4, 0) through (0, 2) and (negative 5, 3).
  5.  

    The graph of y = f of x plots a step function with a set of closed points at (1, 1), (2, 2), and (3, 3) and their corresponding open points at (2, 1), (3, 2), and (4, 3) respectively.

Solution

For the graph of each function, the domain is highlighted in purple on the x-axis and the range is highlighted in green on the y-axis.

  1.  

    The graph of y = f of x is a line that rises from a closed point at (negative 2, 0) to (0, 1) and then rises to (1, 3).
  2.  

    The graph of y = f of x is a line that rises from an open point at (negative 3, 1) to a closed point (2, 2).
  3.  

    The graph of y = f of x is a curve that falls from a closed point at (negative 2, 5) through (negative 1, 2) to (0, 1), and rises to an open point at (1, 2).
  4.  

    The graph of y = f of x is a curve that falls through (negative 5, 3), (0, 2), to a closed point at (4, 0)
  5.  

    The graph of y = f of x plots a step function with a set of closed points at (1, 1), (2, 2), and (3, 3) and their corresponding open points at (2, 1), (3, 2), and (4, 3) respectively.

Check Point 8

Objective 8: Identify the domain and range of a function from its graph

Identifying Domain and Range from a Function’s Graph

  1. Objective 8 Identify the domain and range of a function from its graph.

Watch Video

Figure 1.25 illustrates how the graph of a function is used to determine the function’s domain and its range.

The domain colon set of inputs is labeled, found on the x axis. The range colon set of outputs is labeled, found on the y axis.

Figure 1.25 Domain and range of f

The graph shows a concave down curve.

Figure 1.25 Full Alternative Text

Let’s apply these ideas to the graph of the function shown in Figure 1.26. To find the domain, look for all the inputs on the x-axis that correspond to points on the graph. Can you see that they extend from 4 to 2, inclusive? The function’s domain can be represented as follows:

Examples of using set builder notation and interval notation.
1.2-73 Full Alternative Text

Figure 1.26 Domain and range of f

The graph of y = f of x is a semi circle that rises from (negative 4, 1) to (negative 1, 3) and then falls to (2, 1). The two ends of the curve are marked by closed points.

Figure 1.26 Full Alternative Text

To find the range, look for all the outputs on the y-axis that correspond to points on the graph. They extend from 1 to 4, inclusive. The function’s range can be represented as follows:

Examples of using set builder notation and interval notation.
1.2-74 Full Alternative Text

Example 8 Identifying the Domain and Range of a Function from Its Graph

Use the graph of each function to identify its domain and its range.

  1.  

    The graph of y = f of x is a line that rises from the closed point at (negative 2, 0) to (0, 1) and then to the closed point at (1, 3).
  2.  

    The graph of y = f of x is a line that rises from the open point at (negative 3, 1) to the closed point at (2, 2).
  3.  

    The graph of y = f of x is a curve that falls from the closed point at (negative 2, 5) to (0, 1), and then rises to the open point at (1, 2).
  4.  

    The graph of y = f of x is a curve that rises from the closed point at (4, 0) through (0, 2) and (negative 5, 3).
  5.  

    The graph of y = f of x plots a step function with a set of closed points at (1, 1), (2, 2), and (3, 3) and their corresponding open points at (2, 1), (3, 2), and (4, 3) respectively.

Solution

For the graph of each function, the domain is highlighted in purple on the x-axis and the range is highlighted in green on the y-axis.

  1.  

    The graph of y = f of x is a line that rises from a closed point at (negative 2, 0) to (0, 1) and then rises to (1, 3).
  2.  

    The graph of y = f of x is a line that rises from an open point at (negative 3, 1) to a closed point (2, 2).
  3.  

    The graph of y = f of x is a curve that falls from a closed point at (negative 2, 5) through (negative 1, 2) to (0, 1), and rises to an open point at (1, 2).
  4.  

    The graph of y = f of x is a curve that falls through (negative 5, 3), (0, 2), to a closed point at (4, 0)
  5.  

    The graph of y = f of x plots a step function with a set of closed points at (1, 1), (2, 2), and (3, 3) and their corresponding open points at (2, 1), (3, 2), and (4, 3) respectively.

Check Point 8

Objective 9: Identify intercepts from a function’s graph

Identifying Intercepts from a Function’s Graph

  1. Objective 9 Identify intercepts from a function’s graph.

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Figure 1.27 illustrates how we can identify intercepts from a function’s graph. To find the x-intercepts, look for the points at which the graph crosses the x-axis. There are three such points: (2, 0), (3,0), and (5,0). Thus, the x-intercepts are 2, 3, and 5. We express this in function notation by writing f(2)=0, f(3)=0, and f(5)=0. We say that 2, 3, and 5 are the zeros of the function. The zeros of a function f are the x-values for which f(x)=0. Thus, the real zeros are the x-intercepts.

Figure 1.27 Identifying intercepts

The graph of y = f of x is a curve that rises from an open point at (negative 3, negative 4) through (negative 2, 0) and (0, 2) to reach its maximum at (1, 4).

Figure 1.27 Full Alternative Text

To find the y-intercept, look for the point at which the graph crosses the y-axis. This occurs at (0,3). Thus, the y-intercept is 3. We express this in function notation by writing f(0)=3.

By the definition of a function, for each value of x we can have at most one value for y. What does this mean in terms of intercepts? A function can have more than one x-intercept but at most one y-intercept.

1.2: Exercise Set

1.2 Exercise Set

Practice Exercises

In Exercises 110, determine whether each relation is a function. Give the domain and range for each relation.

  1. 1. {(1, 2), (3, 4), (5, 5)}

  2. 2. {(4, 5), (6, 7), (8, 8)}

  3. 3. {(3, 4), (3, 5), (4, 4), (4, 5)}

  4. 4. {(5, 6), (5, 7), (6, 6), (6, 7)}

  5. 5. {(3, 2), (5, 2), (7, 1), (4, 9)}

  6. 6. {(10, 4), (2, 4), (1, 1), (5, 6)}

  7. 7. {(3, 3), (2, 2), (1, 1), (0, 0)}

  8. 8. {(7, 7), (5, 5), (3, 3), (0, 0)}

  9. 9. {(1, 4), (1, 5), (1, 6)}

  10. 10. {(4, 1), (5, 1), (6, 1)}

In Exercises 1126, determine whether each equation defines y as a function of x.

  1. 11. x+y=16

  2. 12. x+y=25

  3. 13. x2+y=16

  4. 14. x2+y=25

  5. 15. x2+y2=16

  6. 16. x2+y2=25

  7. 17. x=y2

  8. 18. 4x=y2

  9. 19. y=x+4

  10. 20. y=x+4

  11. 21. x+y3=8

  12. 22. x+y3=27

  13. 23. xy+2y=1

  14. 24. xy5y=1

  15. 25. |x|y=2

  16. 26. |x|y=5

In Exercises 2738, evaluate each function at the given values of the independent variable and simplify.

  1. 27. f(x)=4x+5

    1. f(6)

    2. f(x+1)

    3. f(x)

  2. 28. f(x)=3x+7

    1. f(4)

    2. f(x+1)

    3. f(x)

  3. 29. g(x)=x2+2x+3

    1. g(1)

    2. g(x+5)

    3. g(x)

  4. 30. g(x)=x210x3

    1. g(1)

    2. g(x+2)

    3. g(x)

  5. 31. h(x)=x4x2+1

    1. h(2)

    2. h(1)

    3. h(x)

    4. h(3a)

  6. 32. h(x)=x3x+1

    1. h(3)

    2. h(2)

    3. h(x)

    4. h(3a)

  7. 33. f(r)=r+6+3

    1. f(6)

    2. f(10)

    3. f(x6)

  8. 34. f(r)=25r6

    1. f(16)

    2. f(24)

    3. f(252x)

  9. 35. f(x)=4x21x2

    1. f(2)

    2. f(2)

    3. f(x)

  10. 36. f(x)=4x3+1x3

    1. f(2)

    2. f(2)

    3. f(x)

  11. 37. f(x)=x|x|

    1. f(6)

    2. f(6)

    3. f(r2)

  12. 38. f(x)=|x+3|x+3

    1. f(5)

    2. f(5)

    3. f(9x)

In Exercises 3950, graph the given functions, f and g, in the same rectangular coordinate system. Select integers for x, starting with 2 and ending with 2. Once you have obtained your graphs, describe how the graph of g is related to the graph of f.

  1. 39. f(x)=x, g(x)=x+3

  2. 40. f(x)=x, g(x)=x4

  3. 41. f(x)=2x, g(x)=2x1

  4. 42. f(x)=2x, g(x)=2x+3

  5. 43. f(x)=x2, g(x)=x2+1

  6. 44. f(x)=x2, g(x)=x22

  7. 45. f(x)=|x|, g(x)=|x|2

  8. 46. f(x)=|x|, g(x)=|x|+1

  9. 47. f(x)=x3, g(x)=x3+2

  10. 48. f(x)=x3, g(x)=x31

  11. 49. f(x)=3, g(x)=5

  12. 50. f(x)=1, g(x)=4

In Exercises 5154, graph the given square root functions, f and g, in the same rectangular coordinate system. Use the integer values of x given to the right of each function to obtain ordered pairs. Because only nonnegative numbers have square roots that are real numbers, be sure that each graph appears only for values of x that cause the expression under the radical sign to be greater than or equal to zero. Once you have obtained your graphs, describe how the graph of g is related to the graph of f.

  1. 51. f(x)=x(x=0, 1, 4, 9) and g(x)=x1(x=0, 1, 4, 9)

  2. 52. f(x)=x(x=0, 1, 4, 9) and g(x)=x+2(x=0, 1, 4, 9)

  3. 53. f(x)=x(x=0, 1, 4, 9) and g(x)=x1(x=1, 2, 5, 10)

  4. 54. f(x)=x(x=0, 1, 4, 9) and g(x)=x+2(x=2, 1, 2, 7)

In Exercises 5564, use the vertical line test to identify graphs in which y is a function of x.

  1. 55.

    A graph of a line that falls through the second, first, and fourth quadrants.
  2. 56.

    A graph of a line that falls through the first, second, and fourth quadrants.

  3. 57.

    A graph is a horizontal line that passes through the second and first quadrants and is parallel to the x axis.
  4. 58.

    A graph is a vertical line that passes through the first and fourth quadrants and is parallel to the y axis.
  5. 59.

    A graph plots a horizontal ellipse that is centered at a point on the negative x axis and passes through all four quadrants.
  6. 60.

    A graph plots a vertical ellipse that is centered at the origin and passes through all four quadrants.
  7. 61.

    A graph plots a downward opening parabola that rises through the third and second quadrants with indefinite ends with its vertex along the positive y axis then falls through the first and fourth quadrants.
  8. 62.

    A graph plots a leftward opening parabola with indefinite ends that falls through the second quadrant with its vertex at the origin and then continues to fall through the third quadrant.
  9. 63.

    A graph plots two curves. The first curve rises from a closed point on the negative x axis through the second quadrant. The second curve rises from a closed point on the positive x axis through the first quadrant.
  10. 64.

    A graph plots two curves. The first curve falls from an open point on the negative x axis through the third quadrant. The second curve falls from a closed point on the positive x axis through the fourth quadrant.

In Exercises 6570, use the graph of f to find each indicated function value.

  1. 65. f(2)

    The graph of y = f of x is a sinusoidal curve that starts from a closed point at (negative 5, 0) and ends at another closed point at (5, 0). The amplitude and the time period of the sinusoidal curve are 4, each.
  2. 66. f(2)

  3. 67. f(4)

  4. 68. f(4)

  5. 69. f(3)

  6. 70. f(1)

Use the graph of g to solve Exercises 7176.

  1. 71. Find g(4).

    The graph of y = g of x is a curve that passes through (negative 5, 2) to a closed point at (negative 4, 2). The curve falls through another closed point at (0, 0), a third closed point at (2, negative 2), and passes through (3, negative 2).
  2. 72. Find g(2).

  3. 73. Find g(10).

  4. 74. Find g(10).

  5. 75. For what value of x is g(x)=1?

  6. 76. For what value of x is g(x)=1?

In Exercises 7792, use the graph to determine a. the function’s domain; b. the function’s range; c. the x-intercepts, if any; d. the y-intercept, if any; and e. the missing function values, indicated by question marks, below each graph.

  1. 77.

    A graph of y = f of x is an upward opening parabola that falls from the second quadrant through (negative 3, 0) with its vertex at (negative 1, negative 4).
  2. 78.

    A graph of y = f of x is a downward opening parabola that rises from the third quadrant through (negative 3, 0) with its vertex at (negative 1, 4).
  3. 79.

    A graph of y = f of x includes a line that falls through (negative 3, 3) to (0, 1) and another line that rises from (0, 1) through (3, 4). All values are estimated. f of negative 1 and f of 3 are unknown.
  4. 80.

    A graph of y = f of x includes a line that falls through (negative 4, 3) to (negative 1, 0) and another line that rises from (negative 1, 0) through (0, 1). All values are estimated. f of negative 4 and f of 3 are unknown.
  5. 81.

    The graph of y = f of x is a curve that starts at a closed point at (0, negative 1) and rises through (2, 0) to an open point at (5, 5). All values are estimated. f of 3 is unknown.
  6. 82.

    The graph of y = f of x is a curve that starts at an open point at (negative 6, 4) and falls through (negative 4, 0) to a closed point at (0, negative 3). All values are estimated. f of negative 4 is unknown.
  7. 83.

    The graph of y = f of x is a curve that starts at a closed point at (0, 1) and rises through (4, 3). All values are estimated. f of 4 is unknown.
  8. 84.

    The graph of y = f of x is a curve that starts at a closed point at (negative 1, 0) and rises through (0, 1) and (3, 2). All values are estimated. f of 3 is unknown.
  9. 85.

    The graph of y = f of x is a line that falls from a closed point at (negative 2, 6) through (0, 4) and (4, 0) to a closed point at (6, negative 2). All values are estimated. f of negative 1 is unknown.
  10. 86.

    The graph of y = f of x is a line that rises from a closed point at (negative 3, negative 5) through (0, 1) to a closed point at (2, 5). All values are estimated. f of negative 2 is unknown.

  11. 87.

    The graph of y = f of x is a line that rises through (negative 4, negative 5) to (negative 1, negative 2).
  12. 88.

    The graph of y = f of x is a line that passes along the negative x axis till the origin and then rises through (1, 2) and (2, 4). All values are estimated. f of negative 2 and f of 2 are unknown.
  13. 89.

    The graph of y = f of x is concave up decreasing curve with indefinite ends.
  14. 90.

    The graph of y = f of x includes two curves.
  15. 91.

    The graph of y = f of x plots five points as follows, (negative 5, 2), (negative 2, 2), (0, 2), (1, 2), and (3, 2). All values are estimated. f of negative 5 + f of 3 is unknown.
  16. 92.

    The graph of y = f of x plots 5 points at (negative 5, negative 2), (negative 2, negative 2), (0, negative 2), (1, negative 2), and (4, negative 2). All values are estimated. f of negative 5 + f of 4 is unknown.

Practice PLUS

In Exercises 9394, let f(x)=x2x+4 and g(x)=3x5.

  1. 93. Find g(1) and f(g(1)).

  2. 94. Find g(1) and f(g(1)).

In Exercises 9596, let f and g be defined by the following table:

x f(x) g(x)
2 6 0
1 3 4
0 1 1
1 4 3
2 0 6
  1. 95. Find f(1)f(0)[g(2)]2+f(2)÷g(2)g(1).

  2. 96. Find |f(1)f(0)|[g(1)]2+g(1)÷f(1)g(2).

In Exercises 9798, find f(x)f(x) for the given function f. Then simplify the expression.

  1. 97. f(x)=x3+x5

  2. 98. f(x)=x23x+7

Application Exercises

The bar graph shows public spending by the top five and the bottom five countries on pre-primary education and child care. Spending is given by public expenditure as a percentage of gross domestic product. Use the graph to solve Exercises 99100.

A graph titled top five and bottom five countries spending on pre primary education and child care.

Source: USA Today

  1. 99.

    1. Write a set of five ordered pairs in which each of the top-spending countries corresponds to spending as a percentage of gross domestic product. Each ordered pair should be of the form

       (country, percentage of gross domestic product).


    2. Is the relation in part (a) a function? Explain your answer.

    3. For the five top-spending countries, write a set of ordered pairs in which its spending as a percentage of gross domestic product corresponds to the country. Each ordered pair should be of the form

       (percentage of gross domestic product, country).


    4. Is the relation in part (c) a function? Explain your answer.

  2. 100.

    1. Write a set of five ordered pairs in which each of the bottom-spending countries corresponds to spending as a percentage of gross domestic product. Each ordered pair should be of the form

       (country, percentage of gross domestic product).


    2. Is the relation in part (a) a function? Explain your answer.

    3. For the five bottom-spending countries, write a set of ordered pairs in which its spending as a percentage of gross domestic product corresponds to the country. Each ordered pair should be of the form

       (percentage of gross domestic product, country).


    4. Is the relation in part (c) a function? Explain your answer.

The bar graph shows your chances of surviving to various ages once you reach 60.

A graph titled chances of 60 year olds surviving to various ages.

Source: National Center for Health Statistics

The functions

f(x)=2.9x+286andg(x)=0.01x24.9x+370.

model the chance, as a percent, that a 60-year-old will survive to age x. Use this information to solve Exercises 101102.

  1. 101.

    1. Find and interpret f(70).

    2. Find and interpret g(70).

    3. Which function serves as a better model for the chance of surviving to age 70?

  2. 102.

    1. Find and interpret f(90).

    2. Find and interpret g(90).

    3. Which function serves as a better model for the chance of surviving to age 90?

The wage gap, which is used to compare the status of women’s earnings relative to men’s, is expressed as a percent and is calculated by dividing the median, or middlemost, annual earnings for all women by the median annual earnings for all men. The bar graph shows the wage gap for selected years from 1980 through 2020.

A graph titled median women's earnings as a percentage of median men's earnings in the United States.
A graph titled the graph of a function modeling the data, plots wage gap, in percent, versus year after 19 80.

Source: Bureau of Labor Statistics

The function G(x)=0.01x2+0.9x+63 models the wage gap, as a percent, x years after 1980. The graph of function G is shown to the right of the actual data. Use this information to solve Exercises 103104.

  1. 103.

    1. Find and interpret G(40). Identify this information as a point on the graph of the function.

    2. Does G(40) overestimate or underestimate the actual data shown by the bar graph? By how much?

  2. 104.

    1. Find and interpret G(10). Identify this information as a point on the graph of the function.

    2. Does G(10) overestimate or underestimate the actual data shown by the bar graph? By how much?

In Exercises 105108, you will be developing functions that model given conditions.

  1. 105. A company that manufactures bicycles has a fixed cost of $100,000. It costs $100 to produce each bicycle. The total cost for the company is the sum of its fixed cost and variable costs. Write the total cost, C, as a function of the number of bicycles produced, x. Then find and interpret C(90).

  2. 106. A previously owned car was purchased for $22,500. The value of the car decreased by $3200 per year for each of the next six years. Write a function that describes the value of the car, V, after x years, where 0x6. Then find and interpret V(3).

  3. 107. You commute to work a distance of 40 miles and return on the same route at the end of the day. Your average rate on the return trip is 30 miles per hour faster than your average rate on the outgoing trip. Write the total time, T, in hours, devoted to your outgoing and return trips as a function of your rate on the outgoing trip, x. Then find and interpret T(30). Hint:

    Time traveled=Distance traveledRate of travel.
  4. 108. A chemist working on a flu vaccine needs to mix a 10% sodium-iodine solution with a 60% sodium-iodine solution to obtain a 50-milliliter mixture. Write the amount of sodium iodine in the mixture, S, in milliliters, as a function of the number of milliliters of the 10% solution used, x. Then find and interpret S(30).

Explaining the Concepts

  1. 109. What is a relation? Describe what is meant by its domain and its range.

  2. 110. Explain how to determine whether a relation is a function. What is a function?

  3. 111. How do you determine if an equation in x and y defines y as a function of x?

  4. 112. Does f(x) mean f times x when referring to a function f? If not, what does f(x) mean? Provide an example with your explanation.

  5. 113. What is the graph of a function?

  6. 114. Explain how the vertical line test is used to determine whether a graph represents a function.

  7. 115. Explain how to identify the domain and range of a function from its graph.

  8. 116. For people filing a single return, federal income tax is a function of adjusted gross income because for each value of adjusted gross income there is a specific tax to be paid. By contrast, the price of a house is not a function of the lot size on which the house sits because houses on same-sized lots can sell for many different prices.

    1. Describe an everyday situation between variables that is a function.

    2. Describe an everyday situation between variables that is not a function.

Technology Exercise

  1. 117. Use a graphing utility to verify any five pairs of graphs that you drew by hand in Exercises 3954.

Critical Thinking Exercises

MAKE SENSE? In Exercises 118121, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 118. My body temperature is a function of the time of day.

  2. 119. Using f(x)=3x+2, I found f(50) by applying the distributive property to (3x+2)50.

  3. 120. I graphed a function showing how paid vacation days depend on the number of years a person works for a company. The domain was the number of paid vacation days.

  4. 121. I graphed a function showing how the average number of annual physician visits depends on a person’s age. The domain was the average number of annual physician visits.

Use the graph of f to determine whether each statement in Exercises 122125 is true or false.

The graph of f plots a curve and a line.
  1. 122. The domain of f is [4, 1)(1, 4].

  2. 123. The range of f is [2, 2].

  3. 124. f(1)f(4)=2

  4. 125. f(0)=2.1

  5. 126. If f(x)=3x+7, find f(a+h)f(a)h.

  6. 127. Give an example of a relation with the following characteristics: The relation is a function containing two ordered pairs. Reversing the components in each ordered pair results in a relation that is not a function.

  7. 128. If f(x+y)=f(x)+f(y) and f(1)=3, find f(2), f(3), and f(4). Is f(x+y)=f(x)+f(y) for all functions?

Retaining the Concepts

  1. 129. Solve and check: 1 + 3(x4) = 2x. (Section P.7, Example 1)

  2. 130. Solve and check: x35x42=5.

    (Section P.7, Example 2)

  3. 131. Sharks may be scary, but they were responsible for only three deaths worldwide in 2014. The world’s deadliest creatures, ranked by the number of human deaths per year, are mosquitoes, snails, and snakes. The number of deaths by mosquitoes exceeds the number of deaths by snakes by 661 thousand. The number of deaths by snails exceeds the number of deaths by snakes by 106 thousand. Combined, mosquitoes, snails, and snakes result in 1049 thousand (or 1,049,000) human deaths per year. Determine the number of human deaths per year, in thousands, by snakes, mosquitoes, and snails. (Source: World Health Organization) (Section P.8, Example 1)

Preview Exercises

Exercises 132134 will help you prepare for the material covered in the next section.

  1. 132. The function C(g)=40+0.60(g  15) describes the monthly cost, C(g), in dollars, for a high-speed wireless Internet plan for g gigabytes of data, where g>15. Find and interpret C(45).

  2. 133. Use point plotting to graph f(x)=x+2 if x1.

  3. 134. Simplify: 2(x+h)2+3(x+h)+5(2x2+3x+5).

1.2: Exercise Set

1.2 Exercise Set

Practice Exercises

In Exercises 110, determine whether each relation is a function. Give the domain and range for each relation.

  1. 1. {(1, 2), (3, 4), (5, 5)}

  2. 2. {(4, 5), (6, 7), (8, 8)}

  3. 3. {(3, 4), (3, 5), (4, 4), (4, 5)}

  4. 4. {(5, 6), (5, 7), (6, 6), (6, 7)}

  5. 5. {(3, 2), (5, 2), (7, 1), (4, 9)}

  6. 6. {(10, 4), (2, 4), (1, 1), (5, 6)}

  7. 7. {(3, 3), (2, 2), (1, 1), (0, 0)}

  8. 8. {(7, 7), (5, 5), (3, 3), (0, 0)}

  9. 9. {(1, 4), (1, 5), (1, 6)}

  10. 10. {(4, 1), (5, 1), (6, 1)}

In Exercises 1126, determine whether each equation defines y as a function of x.

  1. 11. x+y=16

  2. 12. x+y=25

  3. 13. x2+y=16

  4. 14. x2+y=25

  5. 15. x2+y2=16

  6. 16. x2+y2=25

  7. 17. x=y2

  8. 18. 4x=y2

  9. 19. y=x+4

  10. 20. y=x+4

  11. 21. x+y3=8

  12. 22. x+y3=27

  13. 23. xy+2y=1

  14. 24. xy5y=1

  15. 25. |x|y=2

  16. 26. |x|y=5

In Exercises 2738, evaluate each function at the given values of the independent variable and simplify.

  1. 27. f(x)=4x+5

    1. f(6)

    2. f(x+1)

    3. f(x)

  2. 28. f(x)=3x+7

    1. f(4)

    2. f(x+1)

    3. f(x)

  3. 29. g(x)=x2+2x+3

    1. g(1)

    2. g(x+5)

    3. g(x)

  4. 30. g(x)=x210x3

    1. g(1)

    2. g(x+2)

    3. g(x)

  5. 31. h(x)=x4x2+1

    1. h(2)

    2. h(1)

    3. h(x)

    4. h(3a)

  6. 32. h(x)=x3x+1

    1. h(3)

    2. h(2)

    3. h(x)

    4. h(3a)

  7. 33. f(r)=r+6+3

    1. f(6)

    2. f(10)

    3. f(x6)

  8. 34. f(r)=25r6

    1. f(16)

    2. f(24)

    3. f(252x)

  9. 35. f(x)=4x21x2

    1. f(2)

    2. f(2)

    3. f(x)

  10. 36. f(x)=4x3+1x3

    1. f(2)

    2. f(2)

    3. f(x)

  11. 37. f(x)=x|x|

    1. f(6)

    2. f(6)

    3. f(r2)

  12. 38. f(x)=|x+3|x+3

    1. f(5)

    2. f(5)

    3. f(9x)

In Exercises 3950, graph the given functions, f and g, in the same rectangular coordinate system. Select integers for x, starting with 2 and ending with 2. Once you have obtained your graphs, describe how the graph of g is related to the graph of f.

  1. 39. f(x)=x, g(x)=x+3

  2. 40. f(x)=x, g(x)=x4

  3. 41. f(x)=2x, g(x)=2x1

  4. 42. f(x)=2x, g(x)=2x+3

  5. 43. f(x)=x2, g(x)=x2+1

  6. 44. f(x)=x2, g(x)=x22

  7. 45. f(x)=|x|, g(x)=|x|2

  8. 46. f(x)=|x|, g(x)=|x|+1

  9. 47. f(x)=x3, g(x)=x3+2

  10. 48. f(x)=x3, g(x)=x31

  11. 49. f(x)=3, g(x)=5

  12. 50. f(x)=1, g(x)=4

In Exercises 5154, graph the given square root functions, f and g, in the same rectangular coordinate system. Use the integer values of x given to the right of each function to obtain ordered pairs. Because only nonnegative numbers have square roots that are real numbers, be sure that each graph appears only for values of x that cause the expression under the radical sign to be greater than or equal to zero. Once you have obtained your graphs, describe how the graph of g is related to the graph of f.

  1. 51. f(x)=x(x=0, 1, 4, 9) and g(x)=x1(x=0, 1, 4, 9)

  2. 52. f(x)=x(x=0, 1, 4, 9) and g(x)=x+2(x=0, 1, 4, 9)

  3. 53. f(x)=x(x=0, 1, 4, 9) and g(x)=x1(x=1, 2, 5, 10)

  4. 54. f(x)=x(x=0, 1, 4, 9) and g(x)=x+2(x=2, 1, 2, 7)

In Exercises 5564, use the vertical line test to identify graphs in which y is a function of x.

  1. 55.

    A graph of a line that falls through the second, first, and fourth quadrants.
  2. 56.

    A graph of a line that falls through the first, second, and fourth quadrants.

  3. 57.

    A graph is a horizontal line that passes through the second and first quadrants and is parallel to the x axis.
  4. 58.

    A graph is a vertical line that passes through the first and fourth quadrants and is parallel to the y axis.
  5. 59.

    A graph plots a horizontal ellipse that is centered at a point on the negative x axis and passes through all four quadrants.
  6. 60.

    A graph plots a vertical ellipse that is centered at the origin and passes through all four quadrants.
  7. 61.

    A graph plots a downward opening parabola that rises through the third and second quadrants with indefinite ends with its vertex along the positive y axis then falls through the first and fourth quadrants.
  8. 62.

    A graph plots a leftward opening parabola with indefinite ends that falls through the second quadrant with its vertex at the origin and then continues to fall through the third quadrant.
  9. 63.

    A graph plots two curves. The first curve rises from a closed point on the negative x axis through the second quadrant. The second curve rises from a closed point on the positive x axis through the first quadrant.
  10. 64.

    A graph plots two curves. The first curve falls from an open point on the negative x axis through the third quadrant. The second curve falls from a closed point on the positive x axis through the fourth quadrant.

In Exercises 6570, use the graph of f to find each indicated function value.

  1. 65. f(2)

    The graph of y = f of x is a sinusoidal curve that starts from a closed point at (negative 5, 0) and ends at another closed point at (5, 0). The amplitude and the time period of the sinusoidal curve are 4, each.
  2. 66. f(2)

  3. 67. f(4)

  4. 68. f(4)

  5. 69. f(3)

  6. 70. f(1)

Use the graph of g to solve Exercises 7176.

  1. 71. Find g(4).

    The graph of y = g of x is a curve that passes through (negative 5, 2) to a closed point at (negative 4, 2). The curve falls through another closed point at (0, 0), a third closed point at (2, negative 2), and passes through (3, negative 2).
  2. 72. Find g(2).

  3. 73. Find g(10).

  4. 74. Find g(10).

  5. 75. For what value of x is g(x)=1?

  6. 76. For what value of x is g(x)=1?

In Exercises 7792, use the graph to determine a. the function’s domain; b. the function’s range; c. the x-intercepts, if any; d. the y-intercept, if any; and e. the missing function values, indicated by question marks, below each graph.

  1. 77.

    A graph of y = f of x is an upward opening parabola that falls from the second quadrant through (negative 3, 0) with its vertex at (negative 1, negative 4).
  2. 78.

    A graph of y = f of x is a downward opening parabola that rises from the third quadrant through (negative 3, 0) with its vertex at (negative 1, 4).
  3. 79.

    A graph of y = f of x includes a line that falls through (negative 3, 3) to (0, 1) and another line that rises from (0, 1) through (3, 4). All values are estimated. f of negative 1 and f of 3 are unknown.
  4. 80.

    A graph of y = f of x includes a line that falls through (negative 4, 3) to (negative 1, 0) and another line that rises from (negative 1, 0) through (0, 1). All values are estimated. f of negative 4 and f of 3 are unknown.
  5. 81.

    The graph of y = f of x is a curve that starts at a closed point at (0, negative 1) and rises through (2, 0) to an open point at (5, 5). All values are estimated. f of 3 is unknown.
  6. 82.

    The graph of y = f of x is a curve that starts at an open point at (negative 6, 4) and falls through (negative 4, 0) to a closed point at (0, negative 3). All values are estimated. f of negative 4 is unknown.
  7. 83.

    The graph of y = f of x is a curve that starts at a closed point at (0, 1) and rises through (4, 3). All values are estimated. f of 4 is unknown.
  8. 84.

    The graph of y = f of x is a curve that starts at a closed point at (negative 1, 0) and rises through (0, 1) and (3, 2). All values are estimated. f of 3 is unknown.
  9. 85.

    The graph of y = f of x is a line that falls from a closed point at (negative 2, 6) through (0, 4) and (4, 0) to a closed point at (6, negative 2). All values are estimated. f of negative 1 is unknown.
  10. 86.

    The graph of y = f of x is a line that rises from a closed point at (negative 3, negative 5) through (0, 1) to a closed point at (2, 5). All values are estimated. f of negative 2 is unknown.

  11. 87.

    The graph of y = f of x is a line that rises through (negative 4, negative 5) to (negative 1, negative 2).
  12. 88.

    The graph of y = f of x is a line that passes along the negative x axis till the origin and then rises through (1, 2) and (2, 4). All values are estimated. f of negative 2 and f of 2 are unknown.
  13. 89.

    The graph of y = f of x is concave up decreasing curve with indefinite ends.
  14. 90.

    The graph of y = f of x includes two curves.
  15. 91.

    The graph of y = f of x plots five points as follows, (negative 5, 2), (negative 2, 2), (0, 2), (1, 2), and (3, 2). All values are estimated. f of negative 5 + f of 3 is unknown.
  16. 92.

    The graph of y = f of x plots 5 points at (negative 5, negative 2), (negative 2, negative 2), (0, negative 2), (1, negative 2), and (4, negative 2). All values are estimated. f of negative 5 + f of 4 is unknown.

Practice PLUS

In Exercises 9394, let f(x)=x2x+4 and g(x)=3x5.

  1. 93. Find g(1) and f(g(1)).

  2. 94. Find g(1) and f(g(1)).

In Exercises 9596, let f and g be defined by the following table:

x f(x) g(x)
2 6 0
1 3 4
0 1 1
1 4 3
2 0 6
  1. 95. Find f(1)f(0)[g(2)]2+f(2)÷g(2)g(1).

  2. 96. Find |f(1)f(0)|[g(1)]2+g(1)÷f(1)g(2).

In Exercises 9798, find f(x)f(x) for the given function f. Then simplify the expression.

  1. 97. f(x)=x3+x5

  2. 98. f(x)=x23x+7

Application Exercises

The bar graph shows public spending by the top five and the bottom five countries on pre-primary education and child care. Spending is given by public expenditure as a percentage of gross domestic product. Use the graph to solve Exercises 99100.

A graph titled top five and bottom five countries spending on pre primary education and child care.

Source: USA Today

  1. 99.

    1. Write a set of five ordered pairs in which each of the top-spending countries corresponds to spending as a percentage of gross domestic product. Each ordered pair should be of the form

       (country, percentage of gross domestic product).


    2. Is the relation in part (a) a function? Explain your answer.

    3. For the five top-spending countries, write a set of ordered pairs in which its spending as a percentage of gross domestic product corresponds to the country. Each ordered pair should be of the form

       (percentage of gross domestic product, country).


    4. Is the relation in part (c) a function? Explain your answer.

  2. 100.

    1. Write a set of five ordered pairs in which each of the bottom-spending countries corresponds to spending as a percentage of gross domestic product. Each ordered pair should be of the form

       (country, percentage of gross domestic product).


    2. Is the relation in part (a) a function? Explain your answer.

    3. For the five bottom-spending countries, write a set of ordered pairs in which its spending as a percentage of gross domestic product corresponds to the country. Each ordered pair should be of the form

       (percentage of gross domestic product, country).


    4. Is the relation in part (c) a function? Explain your answer.

The bar graph shows your chances of surviving to various ages once you reach 60.

A graph titled chances of 60 year olds surviving to various ages.

Source: National Center for Health Statistics

The functions

f(x)=2.9x+286andg(x)=0.01x24.9x+370.

model the chance, as a percent, that a 60-year-old will survive to age x. Use this information to solve Exercises 101102.

  1. 101.

    1. Find and interpret f(70).

    2. Find and interpret g(70).

    3. Which function serves as a better model for the chance of surviving to age 70?

  2. 102.

    1. Find and interpret f(90).

    2. Find and interpret g(90).

    3. Which function serves as a better model for the chance of surviving to age 90?

The wage gap, which is used to compare the status of women’s earnings relative to men’s, is expressed as a percent and is calculated by dividing the median, or middlemost, annual earnings for all women by the median annual earnings for all men. The bar graph shows the wage gap for selected years from 1980 through 2020.

A graph titled median women's earnings as a percentage of median men's earnings in the United States.
A graph titled the graph of a function modeling the data, plots wage gap, in percent, versus year after 19 80.

Source: Bureau of Labor Statistics

The function G(x)=0.01x2+0.9x+63 models the wage gap, as a percent, x years after 1980. The graph of function G is shown to the right of the actual data. Use this information to solve Exercises 103104.

  1. 103.

    1. Find and interpret G(40). Identify this information as a point on the graph of the function.

    2. Does G(40) overestimate or underestimate the actual data shown by the bar graph? By how much?

  2. 104.

    1. Find and interpret G(10). Identify this information as a point on the graph of the function.

    2. Does G(10) overestimate or underestimate the actual data shown by the bar graph? By how much?

In Exercises 105108, you will be developing functions that model given conditions.

  1. 105. A company that manufactures bicycles has a fixed cost of $100,000. It costs $100 to produce each bicycle. The total cost for the company is the sum of its fixed cost and variable costs. Write the total cost, C, as a function of the number of bicycles produced, x. Then find and interpret C(90).

  2. 106. A previously owned car was purchased for $22,500. The value of the car decreased by $3200 per year for each of the next six years. Write a function that describes the value of the car, V, after x years, where 0x6. Then find and interpret V(3).

  3. 107. You commute to work a distance of 40 miles and return on the same route at the end of the day. Your average rate on the return trip is 30 miles per hour faster than your average rate on the outgoing trip. Write the total time, T, in hours, devoted to your outgoing and return trips as a function of your rate on the outgoing trip, x. Then find and interpret T(30). Hint:

    Time traveled=Distance traveledRate of travel.
  4. 108. A chemist working on a flu vaccine needs to mix a 10% sodium-iodine solution with a 60% sodium-iodine solution to obtain a 50-milliliter mixture. Write the amount of sodium iodine in the mixture, S, in milliliters, as a function of the number of milliliters of the 10% solution used, x. Then find and interpret S(30).

Explaining the Concepts

  1. 109. What is a relation? Describe what is meant by its domain and its range.

  2. 110. Explain how to determine whether a relation is a function. What is a function?

  3. 111. How do you determine if an equation in x and y defines y as a function of x?

  4. 112. Does f(x) mean f times x when referring to a function f? If not, what does f(x) mean? Provide an example with your explanation.

  5. 113. What is the graph of a function?

  6. 114. Explain how the vertical line test is used to determine whether a graph represents a function.

  7. 115. Explain how to identify the domain and range of a function from its graph.

  8. 116. For people filing a single return, federal income tax is a function of adjusted gross income because for each value of adjusted gross income there is a specific tax to be paid. By contrast, the price of a house is not a function of the lot size on which the house sits because houses on same-sized lots can sell for many different prices.

    1. Describe an everyday situation between variables that is a function.

    2. Describe an everyday situation between variables that is not a function.

Technology Exercise

  1. 117. Use a graphing utility to verify any five pairs of graphs that you drew by hand in Exercises 3954.

Critical Thinking Exercises

MAKE SENSE? In Exercises 118121, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 118. My body temperature is a function of the time of day.

  2. 119. Using f(x)=3x+2, I found f(50) by applying the distributive property to (3x+2)50.

  3. 120. I graphed a function showing how paid vacation days depend on the number of years a person works for a company. The domain was the number of paid vacation days.

  4. 121. I graphed a function showing how the average number of annual physician visits depends on a person’s age. The domain was the average number of annual physician visits.

Use the graph of f to determine whether each statement in Exercises 122125 is true or false.

The graph of f plots a curve and a line.
  1. 122. The domain of f is [4, 1)(1, 4].

  2. 123. The range of f is [2, 2].

  3. 124. f(1)f(4)=2

  4. 125. f(0)=2.1

  5. 126. If f(x)=3x+7, find f(a+h)f(a)h.

  6. 127. Give an example of a relation with the following characteristics: The relation is a function containing two ordered pairs. Reversing the components in each ordered pair results in a relation that is not a function.

  7. 128. If f(x+y)=f(x)+f(y) and f(1)=3, find f(2), f(3), and f(4). Is f(x+y)=f(x)+f(y) for all functions?

Retaining the Concepts

  1. 129. Solve and check: 1 + 3(x4) = 2x. (Section P.7, Example 1)

  2. 130. Solve and check: x35x42=5.

    (Section P.7, Example 2)

  3. 131. Sharks may be scary, but they were responsible for only three deaths worldwide in 2014. The world’s deadliest creatures, ranked by the number of human deaths per year, are mosquitoes, snails, and snakes. The number of deaths by mosquitoes exceeds the number of deaths by snakes by 661 thousand. The number of deaths by snails exceeds the number of deaths by snakes by 106 thousand. Combined, mosquitoes, snails, and snakes result in 1049 thousand (or 1,049,000) human deaths per year. Determine the number of human deaths per year, in thousands, by snakes, mosquitoes, and snails. (Source: World Health Organization) (Section P.8, Example 1)

Preview Exercises

Exercises 132134 will help you prepare for the material covered in the next section.

  1. 132. The function C(g)=40+0.60(g  15) describes the monthly cost, C(g), in dollars, for a high-speed wireless Internet plan for g gigabytes of data, where g>15. Find and interpret C(45).

  2. 133. Use point plotting to graph f(x)=x+2 if x1.

  3. 134. Simplify: 2(x+h)2+3(x+h)+5(2x2+3x+5).

1.2: Exercise Set

1.2 Exercise Set

Practice Exercises

In Exercises 110, determine whether each relation is a function. Give the domain and range for each relation.

  1. 1. {(1, 2), (3, 4), (5, 5)}

  2. 2. {(4, 5), (6, 7), (8, 8)}

  3. 3. {(3, 4), (3, 5), (4, 4), (4, 5)}

  4. 4. {(5, 6), (5, 7), (6, 6), (6, 7)}

  5. 5. {(3, 2), (5, 2), (7, 1), (4, 9)}

  6. 6. {(10, 4), (2, 4), (1, 1), (5, 6)}

  7. 7. {(3, 3), (2, 2), (1, 1), (0, 0)}

  8. 8. {(7, 7), (5, 5), (3, 3), (0, 0)}

  9. 9. {(1, 4), (1, 5), (1, 6)}

  10. 10. {(4, 1), (5, 1), (6, 1)}

In Exercises 1126, determine whether each equation defines y as a function of x.

  1. 11. x+y=16

  2. 12. x+y=25

  3. 13. x2+y=16

  4. 14. x2+y=25

  5. 15. x2+y2=16

  6. 16. x2+y2=25

  7. 17. x=y2

  8. 18. 4x=y2

  9. 19. y=x+4

  10. 20. y=x+4

  11. 21. x+y3=8

  12. 22. x+y3=27

  13. 23. xy+2y=1

  14. 24. xy5y=1

  15. 25. |x|y=2

  16. 26. |x|y=5

In Exercises 2738, evaluate each function at the given values of the independent variable and simplify.

  1. 27. f(x)=4x+5

    1. f(6)

    2. f(x+1)

    3. f(x)

  2. 28. f(x)=3x+7

    1. f(4)

    2. f(x+1)

    3. f(x)

  3. 29. g(x)=x2+2x+3

    1. g(1)

    2. g(x+5)

    3. g(x)

  4. 30. g(x)=x210x3

    1. g(1)

    2. g(x+2)

    3. g(x)

  5. 31. h(x)=x4x2+1

    1. h(2)

    2. h(1)

    3. h(x)

    4. h(3a)

  6. 32. h(x)=x3x+1

    1. h(3)

    2. h(2)

    3. h(x)

    4. h(3a)

  7. 33. f(r)=r+6+3

    1. f(6)

    2. f(10)

    3. f(x6)

  8. 34. f(r)=25r6

    1. f(16)

    2. f(24)

    3. f(252x)

  9. 35. f(x)=4x21x2

    1. f(2)

    2. f(2)

    3. f(x)

  10. 36. f(x)=4x3+1x3

    1. f(2)

    2. f(2)

    3. f(x)

  11. 37. f(x)=x|x|

    1. f(6)

    2. f(6)

    3. f(r2)

  12. 38. f(x)=|x+3|x+3

    1. f(5)

    2. f(5)

    3. f(9x)

In Exercises 3950, graph the given functions, f and g, in the same rectangular coordinate system. Select integers for x, starting with 2 and ending with 2. Once you have obtained your graphs, describe how the graph of g is related to the graph of f.

  1. 39. f(x)=x, g(x)=x+3

  2. 40. f(x)=x, g(x)=x4

  3. 41. f(x)=2x, g(x)=2x1

  4. 42. f(x)=2x, g(x)=2x+3

  5. 43. f(x)=x2, g(x)=x2+1

  6. 44. f(x)=x2, g(x)=x22

  7. 45. f(x)=|x|, g(x)=|x|2

  8. 46. f(x)=|x|, g(x)=|x|+1

  9. 47. f(x)=x3, g(x)=x3+2

  10. 48. f(x)=x3, g(x)=x31

  11. 49. f(x)=3, g(x)=5

  12. 50. f(x)=1, g(x)=4

In Exercises 5154, graph the given square root functions, f and g, in the same rectangular coordinate system. Use the integer values of x given to the right of each function to obtain ordered pairs. Because only nonnegative numbers have square roots that are real numbers, be sure that each graph appears only for values of x that cause the expression under the radical sign to be greater than or equal to zero. Once you have obtained your graphs, describe how the graph of g is related to the graph of f.

  1. 51. f(x)=x(x=0, 1, 4, 9) and g(x)=x1(x=0, 1, 4, 9)

  2. 52. f(x)=x(x=0, 1, 4, 9) and g(x)=x+2(x=0, 1, 4, 9)

  3. 53. f(x)=x(x=0, 1, 4, 9) and g(x)=x1(x=1, 2, 5, 10)

  4. 54. f(x)=x(x=0, 1, 4, 9) and g(x)=x+2(x=2, 1, 2, 7)

In Exercises 5564, use the vertical line test to identify graphs in which y is a function of x.

  1. 55.

    A graph of a line that falls through the second, first, and fourth quadrants.
  2. 56.

    A graph of a line that falls through the first, second, and fourth quadrants.

  3. 57.

    A graph is a horizontal line that passes through the second and first quadrants and is parallel to the x axis.
  4. 58.

    A graph is a vertical line that passes through the first and fourth quadrants and is parallel to the y axis.
  5. 59.

    A graph plots a horizontal ellipse that is centered at a point on the negative x axis and passes through all four quadrants.
  6. 60.

    A graph plots a vertical ellipse that is centered at the origin and passes through all four quadrants.
  7. 61.

    A graph plots a downward opening parabola that rises through the third and second quadrants with indefinite ends with its vertex along the positive y axis then falls through the first and fourth quadrants.
  8. 62.

    A graph plots a leftward opening parabola with indefinite ends that falls through the second quadrant with its vertex at the origin and then continues to fall through the third quadrant.
  9. 63.

    A graph plots two curves. The first curve rises from a closed point on the negative x axis through the second quadrant. The second curve rises from a closed point on the positive x axis through the first quadrant.
  10. 64.

    A graph plots two curves. The first curve falls from an open point on the negative x axis through the third quadrant. The second curve falls from a closed point on the positive x axis through the fourth quadrant.

In Exercises 6570, use the graph of f to find each indicated function value.

  1. 65. f(2)

    The graph of y = f of x is a sinusoidal curve that starts from a closed point at (negative 5, 0) and ends at another closed point at (5, 0). The amplitude and the time period of the sinusoidal curve are 4, each.
  2. 66. f(2)

  3. 67. f(4)

  4. 68. f(4)

  5. 69. f(3)

  6. 70. f(1)

Use the graph of g to solve Exercises 7176.

  1. 71. Find g(4).

    The graph of y = g of x is a curve that passes through (negative 5, 2) to a closed point at (negative 4, 2). The curve falls through another closed point at (0, 0), a third closed point at (2, negative 2), and passes through (3, negative 2).
  2. 72. Find g(2).

  3. 73. Find g(10).

  4. 74. Find g(10).

  5. 75. For what value of x is g(x)=1?

  6. 76. For what value of x is g(x)=1?

In Exercises 7792, use the graph to determine a. the function’s domain; b. the function’s range; c. the x-intercepts, if any; d. the y-intercept, if any; and e. the missing function values, indicated by question marks, below each graph.

  1. 77.

    A graph of y = f of x is an upward opening parabola that falls from the second quadrant through (negative 3, 0) with its vertex at (negative 1, negative 4).
  2. 78.

    A graph of y = f of x is a downward opening parabola that rises from the third quadrant through (negative 3, 0) with its vertex at (negative 1, 4).
  3. 79.

    A graph of y = f of x includes a line that falls through (negative 3, 3) to (0, 1) and another line that rises from (0, 1) through (3, 4). All values are estimated. f of negative 1 and f of 3 are unknown.
  4. 80.

    A graph of y = f of x includes a line that falls through (negative 4, 3) to (negative 1, 0) and another line that rises from (negative 1, 0) through (0, 1). All values are estimated. f of negative 4 and f of 3 are unknown.
  5. 81.

    The graph of y = f of x is a curve that starts at a closed point at (0, negative 1) and rises through (2, 0) to an open point at (5, 5). All values are estimated. f of 3 is unknown.
  6. 82.

    The graph of y = f of x is a curve that starts at an open point at (negative 6, 4) and falls through (negative 4, 0) to a closed point at (0, negative 3). All values are estimated. f of negative 4 is unknown.
  7. 83.

    The graph of y = f of x is a curve that starts at a closed point at (0, 1) and rises through (4, 3). All values are estimated. f of 4 is unknown.
  8. 84.

    The graph of y = f of x is a curve that starts at a closed point at (negative 1, 0) and rises through (0, 1) and (3, 2). All values are estimated. f of 3 is unknown.
  9. 85.

    The graph of y = f of x is a line that falls from a closed point at (negative 2, 6) through (0, 4) and (4, 0) to a closed point at (6, negative 2). All values are estimated. f of negative 1 is unknown.
  10. 86.

    The graph of y = f of x is a line that rises from a closed point at (negative 3, negative 5) through (0, 1) to a closed point at (2, 5). All values are estimated. f of negative 2 is unknown.

  11. 87.

    The graph of y = f of x is a line that rises through (negative 4, negative 5) to (negative 1, negative 2).
  12. 88.

    The graph of y = f of x is a line that passes along the negative x axis till the origin and then rises through (1, 2) and (2, 4). All values are estimated. f of negative 2 and f of 2 are unknown.
  13. 89.

    The graph of y = f of x is concave up decreasing curve with indefinite ends.
  14. 90.

    The graph of y = f of x includes two curves.
  15. 91.

    The graph of y = f of x plots five points as follows, (negative 5, 2), (negative 2, 2), (0, 2), (1, 2), and (3, 2). All values are estimated. f of negative 5 + f of 3 is unknown.
  16. 92.

    The graph of y = f of x plots 5 points at (negative 5, negative 2), (negative 2, negative 2), (0, negative 2), (1, negative 2), and (4, negative 2). All values are estimated. f of negative 5 + f of 4 is unknown.

Practice PLUS

In Exercises 9394, let f(x)=x2x+4 and g(x)=3x5.

  1. 93. Find g(1) and f(g(1)).

  2. 94. Find g(1) and f(g(1)).

In Exercises 9596, let f and g be defined by the following table:

x f(x) g(x)
2 6 0
1 3 4
0 1 1
1 4 3
2 0 6
  1. 95. Find f(1)f(0)[g(2)]2+f(2)÷g(2)g(1).

  2. 96. Find |f(1)f(0)|[g(1)]2+g(1)÷f(1)g(2).

In Exercises 9798, find f(x)f(x) for the given function f. Then simplify the expression.

  1. 97. f(x)=x3+x5

  2. 98. f(x)=x23x+7

Application Exercises

The bar graph shows public spending by the top five and the bottom five countries on pre-primary education and child care. Spending is given by public expenditure as a percentage of gross domestic product. Use the graph to solve Exercises 99100.

A graph titled top five and bottom five countries spending on pre primary education and child care.

Source: USA Today

  1. 99.

    1. Write a set of five ordered pairs in which each of the top-spending countries corresponds to spending as a percentage of gross domestic product. Each ordered pair should be of the form

       (country, percentage of gross domestic product).


    2. Is the relation in part (a) a function? Explain your answer.

    3. For the five top-spending countries, write a set of ordered pairs in which its spending as a percentage of gross domestic product corresponds to the country. Each ordered pair should be of the form

       (percentage of gross domestic product, country).


    4. Is the relation in part (c) a function? Explain your answer.

  2. 100.

    1. Write a set of five ordered pairs in which each of the bottom-spending countries corresponds to spending as a percentage of gross domestic product. Each ordered pair should be of the form

       (country, percentage of gross domestic product).


    2. Is the relation in part (a) a function? Explain your answer.

    3. For the five bottom-spending countries, write a set of ordered pairs in which its spending as a percentage of gross domestic product corresponds to the country. Each ordered pair should be of the form

       (percentage of gross domestic product, country).


    4. Is the relation in part (c) a function? Explain your answer.

The bar graph shows your chances of surviving to various ages once you reach 60.

A graph titled chances of 60 year olds surviving to various ages.

Source: National Center for Health Statistics

The functions

f(x)=2.9x+286andg(x)=0.01x24.9x+370.

model the chance, as a percent, that a 60-year-old will survive to age x. Use this information to solve Exercises 101102.

  1. 101.

    1. Find and interpret f(70).

    2. Find and interpret g(70).

    3. Which function serves as a better model for the chance of surviving to age 70?

  2. 102.

    1. Find and interpret f(90).

    2. Find and interpret g(90).

    3. Which function serves as a better model for the chance of surviving to age 90?

The wage gap, which is used to compare the status of women’s earnings relative to men’s, is expressed as a percent and is calculated by dividing the median, or middlemost, annual earnings for all women by the median annual earnings for all men. The bar graph shows the wage gap for selected years from 1980 through 2020.

A graph titled median women's earnings as a percentage of median men's earnings in the United States.
A graph titled the graph of a function modeling the data, plots wage gap, in percent, versus year after 19 80.

Source: Bureau of Labor Statistics

The function G(x)=0.01x2+0.9x+63 models the wage gap, as a percent, x years after 1980. The graph of function G is shown to the right of the actual data. Use this information to solve Exercises 103104.

  1. 103.

    1. Find and interpret G(40). Identify this information as a point on the graph of the function.

    2. Does G(40) overestimate or underestimate the actual data shown by the bar graph? By how much?

  2. 104.

    1. Find and interpret G(10). Identify this information as a point on the graph of the function.

    2. Does G(10) overestimate or underestimate the actual data shown by the bar graph? By how much?

In Exercises 105108, you will be developing functions that model given conditions.

  1. 105. A company that manufactures bicycles has a fixed cost of $100,000. It costs $100 to produce each bicycle. The total cost for the company is the sum of its fixed cost and variable costs. Write the total cost, C, as a function of the number of bicycles produced, x. Then find and interpret C(90).

  2. 106. A previously owned car was purchased for $22,500. The value of the car decreased by $3200 per year for each of the next six years. Write a function that describes the value of the car, V, after x years, where 0x6. Then find and interpret V(3).

  3. 107. You commute to work a distance of 40 miles and return on the same route at the end of the day. Your average rate on the return trip is 30 miles per hour faster than your average rate on the outgoing trip. Write the total time, T, in hours, devoted to your outgoing and return trips as a function of your rate on the outgoing trip, x. Then find and interpret T(30). Hint:

    Time traveled=Distance traveledRate of travel.
  4. 108. A chemist working on a flu vaccine needs to mix a 10% sodium-iodine solution with a 60% sodium-iodine solution to obtain a 50-milliliter mixture. Write the amount of sodium iodine in the mixture, S, in milliliters, as a function of the number of milliliters of the 10% solution used, x. Then find and interpret S(30).

Explaining the Concepts

  1. 109. What is a relation? Describe what is meant by its domain and its range.

  2. 110. Explain how to determine whether a relation is a function. What is a function?

  3. 111. How do you determine if an equation in x and y defines y as a function of x?

  4. 112. Does f(x) mean f times x when referring to a function f? If not, what does f(x) mean? Provide an example with your explanation.

  5. 113. What is the graph of a function?

  6. 114. Explain how the vertical line test is used to determine whether a graph represents a function.

  7. 115. Explain how to identify the domain and range of a function from its graph.

  8. 116. For people filing a single return, federal income tax is a function of adjusted gross income because for each value of adjusted gross income there is a specific tax to be paid. By contrast, the price of a house is not a function of the lot size on which the house sits because houses on same-sized lots can sell for many different prices.

    1. Describe an everyday situation between variables that is a function.

    2. Describe an everyday situation between variables that is not a function.

Technology Exercise

  1. 117. Use a graphing utility to verify any five pairs of graphs that you drew by hand in Exercises 3954.

Critical Thinking Exercises

MAKE SENSE? In Exercises 118121, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 118. My body temperature is a function of the time of day.

  2. 119. Using f(x)=3x+2, I found f(50) by applying the distributive property to (3x+2)50.

  3. 120. I graphed a function showing how paid vacation days depend on the number of years a person works for a company. The domain was the number of paid vacation days.

  4. 121. I graphed a function showing how the average number of annual physician visits depends on a person’s age. The domain was the average number of annual physician visits.

Use the graph of f to determine whether each statement in Exercises 122125 is true or false.

The graph of f plots a curve and a line.
  1. 122. The domain of f is [4, 1)(1, 4].

  2. 123. The range of f is [2, 2].

  3. 124. f(1)f(4)=2

  4. 125. f(0)=2.1

  5. 126. If f(x)=3x+7, find f(a+h)f(a)h.

  6. 127. Give an example of a relation with the following characteristics: The relation is a function containing two ordered pairs. Reversing the components in each ordered pair results in a relation that is not a function.

  7. 128. If f(x+y)=f(x)+f(y) and f(1)=3, find f(2), f(3), and f(4). Is f(x+y)=f(x)+f(y) for all functions?

Retaining the Concepts

  1. 129. Solve and check: 1 + 3(x4) = 2x. (Section P.7, Example 1)

  2. 130. Solve and check: x35x42=5.

    (Section P.7, Example 2)

  3. 131. Sharks may be scary, but they were responsible for only three deaths worldwide in 2014. The world’s deadliest creatures, ranked by the number of human deaths per year, are mosquitoes, snails, and snakes. The number of deaths by mosquitoes exceeds the number of deaths by snakes by 661 thousand. The number of deaths by snails exceeds the number of deaths by snakes by 106 thousand. Combined, mosquitoes, snails, and snakes result in 1049 thousand (or 1,049,000) human deaths per year. Determine the number of human deaths per year, in thousands, by snakes, mosquitoes, and snails. (Source: World Health Organization) (Section P.8, Example 1)

Preview Exercises

Exercises 132134 will help you prepare for the material covered in the next section.

  1. 132. The function C(g)=40+0.60(g  15) describes the monthly cost, C(g), in dollars, for a high-speed wireless Internet plan for g gigabytes of data, where g>15. Find and interpret C(45).

  2. 133. Use point plotting to graph f(x)=x+2 if x1.

  3. 134. Simplify: 2(x+h)2+3(x+h)+5(2x2+3x+5).

1.2: Exercise Set

1.2 Exercise Set

Practice Exercises

In Exercises 110, determine whether each relation is a function. Give the domain and range for each relation.

  1. 1. {(1, 2), (3, 4), (5, 5)}

  2. 2. {(4, 5), (6, 7), (8, 8)}

  3. 3. {(3, 4), (3, 5), (4, 4), (4, 5)}

  4. 4. {(5, 6), (5, 7), (6, 6), (6, 7)}

  5. 5. {(3, 2), (5, 2), (7, 1), (4, 9)}

  6. 6. {(10, 4), (2, 4), (1, 1), (5, 6)}

  7. 7. {(3, 3), (2, 2), (1, 1), (0, 0)}

  8. 8. {(7, 7), (5, 5), (3, 3), (0, 0)}

  9. 9. {(1, 4), (1, 5), (1, 6)}

  10. 10. {(4, 1), (5, 1), (6, 1)}

In Exercises 1126, determine whether each equation defines y as a function of x.

  1. 11. x+y=16

  2. 12. x+y=25

  3. 13. x2+y=16

  4. 14. x2+y=25

  5. 15. x2+y2=16

  6. 16. x2+y2=25

  7. 17. x=y2

  8. 18. 4x=y2

  9. 19. y=x+4

  10. 20. y=x+4

  11. 21. x+y3=8

  12. 22. x+y3=27

  13. 23. xy+2y=1

  14. 24. xy5y=1

  15. 25. |x|y=2

  16. 26. |x|y=5

In Exercises 2738, evaluate each function at the given values of the independent variable and simplify.

  1. 27. f(x)=4x+5

    1. f(6)

    2. f(x+1)

    3. f(x)

  2. 28. f(x)=3x+7

    1. f(4)

    2. f(x+1)

    3. f(x)

  3. 29. g(x)=x2+2x+3

    1. g(1)

    2. g(x+5)

    3. g(x)

  4. 30. g(x)=x210x3

    1. g(1)

    2. g(x+2)

    3. g(x)

  5. 31. h(x)=x4x2+1

    1. h(2)

    2. h(1)

    3. h(x)

    4. h(3a)

  6. 32. h(x)=x3x+1

    1. h(3)

    2. h(2)

    3. h(x)

    4. h(3a)

  7. 33. f(r)=r+6+3

    1. f(6)

    2. f(10)

    3. f(x6)

  8. 34. f(r)=25r6

    1. f(16)

    2. f(24)

    3. f(252x)

  9. 35. f(x)=4x21x2

    1. f(2)

    2. f(2)

    3. f(x)

  10. 36. f(x)=4x3+1x3

    1. f(2)

    2. f(2)

    3. f(x)

  11. 37. f(x)=x|x|

    1. f(6)

    2. f(6)

    3. f(r2)

  12. 38. f(x)=|x+3|x+3

    1. f(5)

    2. f(5)

    3. f(9x)

In Exercises 3950, graph the given functions, f and g, in the same rectangular coordinate system. Select integers for x, starting with 2 and ending with 2. Once you have obtained your graphs, describe how the graph of g is related to the graph of f.

  1. 39. f(x)=x, g(x)=x+3

  2. 40. f(x)=x, g(x)=x4

  3. 41. f(x)=2x, g(x)=2x1

  4. 42. f(x)=2x, g(x)=2x+3

  5. 43. f(x)=x2, g(x)=x2+1

  6. 44. f(x)=x2, g(x)=x22

  7. 45. f(x)=|x|, g(x)=|x|2

  8. 46. f(x)=|x|, g(x)=|x|+1

  9. 47. f(x)=x3, g(x)=x3+2

  10. 48. f(x)=x3, g(x)=x31

  11. 49. f(x)=3, g(x)=5

  12. 50. f(x)=1, g(x)=4

In Exercises 5154, graph the given square root functions, f and g, in the same rectangular coordinate system. Use the integer values of x given to the right of each function to obtain ordered pairs. Because only nonnegative numbers have square roots that are real numbers, be sure that each graph appears only for values of x that cause the expression under the radical sign to be greater than or equal to zero. Once you have obtained your graphs, describe how the graph of g is related to the graph of f.

  1. 51. f(x)=x(x=0, 1, 4, 9) and g(x)=x1(x=0, 1, 4, 9)

  2. 52. f(x)=x(x=0, 1, 4, 9) and g(x)=x+2(x=0, 1, 4, 9)

  3. 53. f(x)=x(x=0, 1, 4, 9) and g(x)=x1(x=1, 2, 5, 10)

  4. 54. f(x)=x(x=0, 1, 4, 9) and g(x)=x+2(x=2, 1, 2, 7)

In Exercises 5564, use the vertical line test to identify graphs in which y is a function of x.

  1. 55.

    A graph of a line that falls through the second, first, and fourth quadrants.
  2. 56.

    A graph of a line that falls through the first, second, and fourth quadrants.

  3. 57.

    A graph is a horizontal line that passes through the second and first quadrants and is parallel to the x axis.
  4. 58.

    A graph is a vertical line that passes through the first and fourth quadrants and is parallel to the y axis.
  5. 59.

    A graph plots a horizontal ellipse that is centered at a point on the negative x axis and passes through all four quadrants.
  6. 60.

    A graph plots a vertical ellipse that is centered at the origin and passes through all four quadrants.
  7. 61.

    A graph plots a downward opening parabola that rises through the third and second quadrants with indefinite ends with its vertex along the positive y axis then falls through the first and fourth quadrants.
  8. 62.

    A graph plots a leftward opening parabola with indefinite ends that falls through the second quadrant with its vertex at the origin and then continues to fall through the third quadrant.
  9. 63.

    A graph plots two curves. The first curve rises from a closed point on the negative x axis through the second quadrant. The second curve rises from a closed point on the positive x axis through the first quadrant.
  10. 64.

    A graph plots two curves. The first curve falls from an open point on the negative x axis through the third quadrant. The second curve falls from a closed point on the positive x axis through the fourth quadrant.

In Exercises 6570, use the graph of f to find each indicated function value.

  1. 65. f(2)

    The graph of y = f of x is a sinusoidal curve that starts from a closed point at (negative 5, 0) and ends at another closed point at (5, 0). The amplitude and the time period of the sinusoidal curve are 4, each.
  2. 66. f(2)

  3. 67. f(4)

  4. 68. f(4)

  5. 69. f(3)

  6. 70. f(1)

Use the graph of g to solve Exercises 7176.

  1. 71. Find g(4).

    The graph of y = g of x is a curve that passes through (negative 5, 2) to a closed point at (negative 4, 2). The curve falls through another closed point at (0, 0), a third closed point at (2, negative 2), and passes through (3, negative 2).
  2. 72. Find g(2).

  3. 73. Find g(10).

  4. 74. Find g(10).

  5. 75. For what value of x is g(x)=1?

  6. 76. For what value of x is g(x)=1?

In Exercises 7792, use the graph to determine a. the function’s domain; b. the function’s range; c. the x-intercepts, if any; d. the y-intercept, if any; and e. the missing function values, indicated by question marks, below each graph.

  1. 77.

    A graph of y = f of x is an upward opening parabola that falls from the second quadrant through (negative 3, 0) with its vertex at (negative 1, negative 4).
  2. 78.

    A graph of y = f of x is a downward opening parabola that rises from the third quadrant through (negative 3, 0) with its vertex at (negative 1, 4).
  3. 79.

    A graph of y = f of x includes a line that falls through (negative 3, 3) to (0, 1) and another line that rises from (0, 1) through (3, 4). All values are estimated. f of negative 1 and f of 3 are unknown.
  4. 80.

    A graph of y = f of x includes a line that falls through (negative 4, 3) to (negative 1, 0) and another line that rises from (negative 1, 0) through (0, 1). All values are estimated. f of negative 4 and f of 3 are unknown.
  5. 81.

    The graph of y = f of x is a curve that starts at a closed point at (0, negative 1) and rises through (2, 0) to an open point at (5, 5). All values are estimated. f of 3 is unknown.
  6. 82.

    The graph of y = f of x is a curve that starts at an open point at (negative 6, 4) and falls through (negative 4, 0) to a closed point at (0, negative 3). All values are estimated. f of negative 4 is unknown.
  7. 83.

    The graph of y = f of x is a curve that starts at a closed point at (0, 1) and rises through (4, 3). All values are estimated. f of 4 is unknown.
  8. 84.

    The graph of y = f of x is a curve that starts at a closed point at (negative 1, 0) and rises through (0, 1) and (3, 2). All values are estimated. f of 3 is unknown.
  9. 85.

    The graph of y = f of x is a line that falls from a closed point at (negative 2, 6) through (0, 4) and (4, 0) to a closed point at (6, negative 2). All values are estimated. f of negative 1 is unknown.
  10. 86.

    The graph of y = f of x is a line that rises from a closed point at (negative 3, negative 5) through (0, 1) to a closed point at (2, 5). All values are estimated. f of negative 2 is unknown.

  11. 87.

    The graph of y = f of x is a line that rises through (negative 4, negative 5) to (negative 1, negative 2).
  12. 88.

    The graph of y = f of x is a line that passes along the negative x axis till the origin and then rises through (1, 2) and (2, 4). All values are estimated. f of negative 2 and f of 2 are unknown.
  13. 89.

    The graph of y = f of x is concave up decreasing curve with indefinite ends.
  14. 90.

    The graph of y = f of x includes two curves.
  15. 91.

    The graph of y = f of x plots five points as follows, (negative 5, 2), (negative 2, 2), (0, 2), (1, 2), and (3, 2). All values are estimated. f of negative 5 + f of 3 is unknown.
  16. 92.

    The graph of y = f of x plots 5 points at (negative 5, negative 2), (negative 2, negative 2), (0, negative 2), (1, negative 2), and (4, negative 2). All values are estimated. f of negative 5 + f of 4 is unknown.

Practice PLUS

In Exercises 9394, let f(x)=x2x+4 and g(x)=3x5.

  1. 93. Find g(1) and f(g(1)).

  2. 94. Find g(1) and f(g(1)).

In Exercises 9596, let f and g be defined by the following table:

x f(x) g(x)
2 6 0
1 3 4
0 1 1
1 4 3
2 0 6
  1. 95. Find f(1)f(0)[g(2)]2+f(2)÷g(2)g(1).

  2. 96. Find |f(1)f(0)|[g(1)]2+g(1)÷f(1)g(2).

In Exercises 9798, find f(x)f(x) for the given function f. Then simplify the expression.

  1. 97. f(x)=x3+x5

  2. 98. f(x)=x23x+7

Application Exercises

The bar graph shows public spending by the top five and the bottom five countries on pre-primary education and child care. Spending is given by public expenditure as a percentage of gross domestic product. Use the graph to solve Exercises 99100.

A graph titled top five and bottom five countries spending on pre primary education and child care.

Source: USA Today

  1. 99.

    1. Write a set of five ordered pairs in which each of the top-spending countries corresponds to spending as a percentage of gross domestic product. Each ordered pair should be of the form

       (country, percentage of gross domestic product).


    2. Is the relation in part (a) a function? Explain your answer.

    3. For the five top-spending countries, write a set of ordered pairs in which its spending as a percentage of gross domestic product corresponds to the country. Each ordered pair should be of the form

       (percentage of gross domestic product, country).


    4. Is the relation in part (c) a function? Explain your answer.

  2. 100.

    1. Write a set of five ordered pairs in which each of the bottom-spending countries corresponds to spending as a percentage of gross domestic product. Each ordered pair should be of the form

       (country, percentage of gross domestic product).


    2. Is the relation in part (a) a function? Explain your answer.

    3. For the five bottom-spending countries, write a set of ordered pairs in which its spending as a percentage of gross domestic product corresponds to the country. Each ordered pair should be of the form

       (percentage of gross domestic product, country).


    4. Is the relation in part (c) a function? Explain your answer.

The bar graph shows your chances of surviving to various ages once you reach 60.

A graph titled chances of 60 year olds surviving to various ages.

Source: National Center for Health Statistics

The functions

f(x)=2.9x+286andg(x)=0.01x24.9x+370.

model the chance, as a percent, that a 60-year-old will survive to age x. Use this information to solve Exercises 101102.

  1. 101.

    1. Find and interpret f(70).

    2. Find and interpret g(70).

    3. Which function serves as a better model for the chance of surviving to age 70?

  2. 102.

    1. Find and interpret f(90).

    2. Find and interpret g(90).

    3. Which function serves as a better model for the chance of surviving to age 90?

The wage gap, which is used to compare the status of women’s earnings relative to men’s, is expressed as a percent and is calculated by dividing the median, or middlemost, annual earnings for all women by the median annual earnings for all men. The bar graph shows the wage gap for selected years from 1980 through 2020.

A graph titled median women's earnings as a percentage of median men's earnings in the United States.
A graph titled the graph of a function modeling the data, plots wage gap, in percent, versus year after 19 80.

Source: Bureau of Labor Statistics

The function G(x)=0.01x2+0.9x+63 models the wage gap, as a percent, x years after 1980. The graph of function G is shown to the right of the actual data. Use this information to solve Exercises 103104.

  1. 103.

    1. Find and interpret G(40). Identify this information as a point on the graph of the function.

    2. Does G(40) overestimate or underestimate the actual data shown by the bar graph? By how much?

  2. 104.

    1. Find and interpret G(10). Identify this information as a point on the graph of the function.

    2. Does G(10) overestimate or underestimate the actual data shown by the bar graph? By how much?

In Exercises 105108, you will be developing functions that model given conditions.

  1. 105. A company that manufactures bicycles has a fixed cost of $100,000. It costs $100 to produce each bicycle. The total cost for the company is the sum of its fixed cost and variable costs. Write the total cost, C, as a function of the number of bicycles produced, x. Then find and interpret C(90).

  2. 106. A previously owned car was purchased for $22,500. The value of the car decreased by $3200 per year for each of the next six years. Write a function that describes the value of the car, V, after x years, where 0x6. Then find and interpret V(3).

  3. 107. You commute to work a distance of 40 miles and return on the same route at the end of the day. Your average rate on the return trip is 30 miles per hour faster than your average rate on the outgoing trip. Write the total time, T, in hours, devoted to your outgoing and return trips as a function of your rate on the outgoing trip, x. Then find and interpret T(30). Hint:

    Time traveled=Distance traveledRate of travel.
  4. 108. A chemist working on a flu vaccine needs to mix a 10% sodium-iodine solution with a 60% sodium-iodine solution to obtain a 50-milliliter mixture. Write the amount of sodium iodine in the mixture, S, in milliliters, as a function of the number of milliliters of the 10% solution used, x. Then find and interpret S(30).

Explaining the Concepts

  1. 109. What is a relation? Describe what is meant by its domain and its range.

  2. 110. Explain how to determine whether a relation is a function. What is a function?

  3. 111. How do you determine if an equation in x and y defines y as a function of x?

  4. 112. Does f(x) mean f times x when referring to a function f? If not, what does f(x) mean? Provide an example with your explanation.

  5. 113. What is the graph of a function?

  6. 114. Explain how the vertical line test is used to determine whether a graph represents a function.

  7. 115. Explain how to identify the domain and range of a function from its graph.

  8. 116. For people filing a single return, federal income tax is a function of adjusted gross income because for each value of adjusted gross income there is a specific tax to be paid. By contrast, the price of a house is not a function of the lot size on which the house sits because houses on same-sized lots can sell for many different prices.

    1. Describe an everyday situation between variables that is a function.

    2. Describe an everyday situation between variables that is not a function.

Technology Exercise

  1. 117. Use a graphing utility to verify any five pairs of graphs that you drew by hand in Exercises 3954.

Critical Thinking Exercises

MAKE SENSE? In Exercises 118121, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 118. My body temperature is a function of the time of day.

  2. 119. Using f(x)=3x+2, I found f(50) by applying the distributive property to (3x+2)50.

  3. 120. I graphed a function showing how paid vacation days depend on the number of years a person works for a company. The domain was the number of paid vacation days.

  4. 121. I graphed a function showing how the average number of annual physician visits depends on a person’s age. The domain was the average number of annual physician visits.

Use the graph of f to determine whether each statement in Exercises 122125 is true or false.

The graph of f plots a curve and a line.
  1. 122. The domain of f is [4, 1)(1, 4].

  2. 123. The range of f is [2, 2].

  3. 124. f(1)f(4)=2

  4. 125. f(0)=2.1

  5. 126. If f(x)=3x+7, find f(a+h)f(a)h.

  6. 127. Give an example of a relation with the following characteristics: The relation is a function containing two ordered pairs. Reversing the components in each ordered pair results in a relation that is not a function.

  7. 128. If f(x+y)=f(x)+f(y) and f(1)=3, find f(2), f(3), and f(4). Is f(x+y)=f(x)+f(y) for all functions?

Retaining the Concepts

  1. 129. Solve and check: 1 + 3(x4) = 2x. (Section P.7, Example 1)

  2. 130. Solve and check: x35x42=5.

    (Section P.7, Example 2)

  3. 131. Sharks may be scary, but they were responsible for only three deaths worldwide in 2014. The world’s deadliest creatures, ranked by the number of human deaths per year, are mosquitoes, snails, and snakes. The number of deaths by mosquitoes exceeds the number of deaths by snakes by 661 thousand. The number of deaths by snails exceeds the number of deaths by snakes by 106 thousand. Combined, mosquitoes, snails, and snakes result in 1049 thousand (or 1,049,000) human deaths per year. Determine the number of human deaths per year, in thousands, by snakes, mosquitoes, and snails. (Source: World Health Organization) (Section P.8, Example 1)

Preview Exercises

Exercises 132134 will help you prepare for the material covered in the next section.

  1. 132. The function C(g)=40+0.60(g  15) describes the monthly cost, C(g), in dollars, for a high-speed wireless Internet plan for g gigabytes of data, where g>15. Find and interpret C(45).

  2. 133. Use point plotting to graph f(x)=x+2 if x1.

  3. 134. Simplify: 2(x+h)2+3(x+h)+5(2x2+3x+5).

1.2: Exercise Set

1.2 Exercise Set

Practice Exercises

In Exercises 110, determine whether each relation is a function. Give the domain and range for each relation.

  1. 1. {(1, 2), (3, 4), (5, 5)}

  2. 2. {(4, 5), (6, 7), (8, 8)}

  3. 3. {(3, 4), (3, 5), (4, 4), (4, 5)}

  4. 4. {(5, 6), (5, 7), (6, 6), (6, 7)}

  5. 5. {(3, 2), (5, 2), (7, 1), (4, 9)}

  6. 6. {(10, 4), (2, 4), (1, 1), (5, 6)}

  7. 7. {(3, 3), (2, 2), (1, 1), (0, 0)}

  8. 8. {(7, 7), (5, 5), (3, 3), (0, 0)}

  9. 9. {(1, 4), (1, 5), (1, 6)}

  10. 10. {(4, 1), (5, 1), (6, 1)}

In Exercises 1126, determine whether each equation defines y as a function of x.

  1. 11. x+y=16

  2. 12. x+y=25

  3. 13. x2+y=16

  4. 14. x2+y=25

  5. 15. x2+y2=16

  6. 16. x2+y2=25

  7. 17. x=y2

  8. 18. 4x=y2

  9. 19. y=x+4

  10. 20. y=x+4

  11. 21. x+y3=8

  12. 22. x+y3=27

  13. 23. xy+2y=1

  14. 24. xy5y=1

  15. 25. |x|y=2

  16. 26. |x|y=5

In Exercises 2738, evaluate each function at the given values of the independent variable and simplify.

  1. 27. f(x)=4x+5

    1. f(6)

    2. f(x+1)

    3. f(x)

  2. 28. f(x)=3x+7

    1. f(4)

    2. f(x+1)

    3. f(x)

  3. 29. g(x)=x2+2x+3

    1. g(1)

    2. g(x+5)

    3. g(x)

  4. 30. g(x)=x210x3

    1. g(1)

    2. g(x+2)

    3. g(x)

  5. 31. h(x)=x4x2+1

    1. h(2)

    2. h(1)

    3. h(x)

    4. h(3a)

  6. 32. h(x)=x3x+1

    1. h(3)

    2. h(2)

    3. h(x)

    4. h(3a)

  7. 33. f(r)=r+6+3

    1. f(6)

    2. f(10)

    3. f(x6)

  8. 34. f(r)=25r6

    1. f(16)

    2. f(24)

    3. f(252x)

  9. 35. f(x)=4x21x2

    1. f(2)

    2. f(2)

    3. f(x)

  10. 36. f(x)=4x3+1x3

    1. f(2)

    2. f(2)

    3. f(x)

  11. 37. f(x)=x|x|

    1. f(6)

    2. f(6)

    3. f(r2)

  12. 38. f(x)=|x+3|x+3

    1. f(5)

    2. f(5)

    3. f(9x)

In Exercises 3950, graph the given functions, f and g, in the same rectangular coordinate system. Select integers for x, starting with 2 and ending with 2. Once you have obtained your graphs, describe how the graph of g is related to the graph of f.

  1. 39. f(x)=x, g(x)=x+3

  2. 40. f(x)=x, g(x)=x4

  3. 41. f(x)=2x, g(x)=2x1

  4. 42. f(x)=2x, g(x)=2x+3

  5. 43. f(x)=x2, g(x)=x2+1

  6. 44. f(x)=x2, g(x)=x22

  7. 45. f(x)=|x|, g(x)=|x|2

  8. 46. f(x)=|x|, g(x)=|x|+1

  9. 47. f(x)=x3, g(x)=x3+2

  10. 48. f(x)=x3, g(x)=x31

  11. 49. f(x)=3, g(x)=5

  12. 50. f(x)=1, g(x)=4

In Exercises 5154, graph the given square root functions, f and g, in the same rectangular coordinate system. Use the integer values of x given to the right of each function to obtain ordered pairs. Because only nonnegative numbers have square roots that are real numbers, be sure that each graph appears only for values of x that cause the expression under the radical sign to be greater than or equal to zero. Once you have obtained your graphs, describe how the graph of g is related to the graph of f.

  1. 51. f(x)=x(x=0, 1, 4, 9) and g(x)=x1(x=0, 1, 4, 9)

  2. 52. f(x)=x(x=0, 1, 4, 9) and g(x)=x+2(x=0, 1, 4, 9)

  3. 53. f(x)=x(x=0, 1, 4, 9) and g(x)=x1(x=1, 2, 5, 10)

  4. 54. f(x)=x(x=0, 1, 4, 9) and g(x)=x+2(x=2, 1, 2, 7)

In Exercises 5564, use the vertical line test to identify graphs in which y is a function of x.

  1. 55.

    A graph of a line that falls through the second, first, and fourth quadrants.
  2. 56.

    A graph of a line that falls through the first, second, and fourth quadrants.

  3. 57.

    A graph is a horizontal line that passes through the second and first quadrants and is parallel to the x axis.
  4. 58.

    A graph is a vertical line that passes through the first and fourth quadrants and is parallel to the y axis.
  5. 59.

    A graph plots a horizontal ellipse that is centered at a point on the negative x axis and passes through all four quadrants.
  6. 60.

    A graph plots a vertical ellipse that is centered at the origin and passes through all four quadrants.
  7. 61.

    A graph plots a downward opening parabola that rises through the third and second quadrants with indefinite ends with its vertex along the positive y axis then falls through the first and fourth quadrants.
  8. 62.

    A graph plots a leftward opening parabola with indefinite ends that falls through the second quadrant with its vertex at the origin and then continues to fall through the third quadrant.
  9. 63.

    A graph plots two curves. The first curve rises from a closed point on the negative x axis through the second quadrant. The second curve rises from a closed point on the positive x axis through the first quadrant.
  10. 64.

    A graph plots two curves. The first curve falls from an open point on the negative x axis through the third quadrant. The second curve falls from a closed point on the positive x axis through the fourth quadrant.

In Exercises 6570, use the graph of f to find each indicated function value.

  1. 65. f(2)

    The graph of y = f of x is a sinusoidal curve that starts from a closed point at (negative 5, 0) and ends at another closed point at (5, 0). The amplitude and the time period of the sinusoidal curve are 4, each.
  2. 66. f(2)

  3. 67. f(4)

  4. 68. f(4)

  5. 69. f(3)

  6. 70. f(1)

Use the graph of g to solve Exercises 7176.

  1. 71. Find g(4).

    The graph of y = g of x is a curve that passes through (negative 5, 2) to a closed point at (negative 4, 2). The curve falls through another closed point at (0, 0), a third closed point at (2, negative 2), and passes through (3, negative 2).
  2. 72. Find g(2).

  3. 73. Find g(10).

  4. 74. Find g(10).

  5. 75. For what value of x is g(x)=1?

  6. 76. For what value of x is g(x)=1?

In Exercises 7792, use the graph to determine a. the function’s domain; b. the function’s range; c. the x-intercepts, if any; d. the y-intercept, if any; and e. the missing function values, indicated by question marks, below each graph.

  1. 77.

    A graph of y = f of x is an upward opening parabola that falls from the second quadrant through (negative 3, 0) with its vertex at (negative 1, negative 4).
  2. 78.

    A graph of y = f of x is a downward opening parabola that rises from the third quadrant through (negative 3, 0) with its vertex at (negative 1, 4).
  3. 79.

    A graph of y = f of x includes a line that falls through (negative 3, 3) to (0, 1) and another line that rises from (0, 1) through (3, 4). All values are estimated. f of negative 1 and f of 3 are unknown.
  4. 80.

    A graph of y = f of x includes a line that falls through (negative 4, 3) to (negative 1, 0) and another line that rises from (negative 1, 0) through (0, 1). All values are estimated. f of negative 4 and f of 3 are unknown.
  5. 81.

    The graph of y = f of x is a curve that starts at a closed point at (0, negative 1) and rises through (2, 0) to an open point at (5, 5). All values are estimated. f of 3 is unknown.
  6. 82.

    The graph of y = f of x is a curve that starts at an open point at (negative 6, 4) and falls through (negative 4, 0) to a closed point at (0, negative 3). All values are estimated. f of negative 4 is unknown.
  7. 83.

    The graph of y = f of x is a curve that starts at a closed point at (0, 1) and rises through (4, 3). All values are estimated. f of 4 is unknown.
  8. 84.

    The graph of y = f of x is a curve that starts at a closed point at (negative 1, 0) and rises through (0, 1) and (3, 2). All values are estimated. f of 3 is unknown.
  9. 85.

    The graph of y = f of x is a line that falls from a closed point at (negative 2, 6) through (0, 4) and (4, 0) to a closed point at (6, negative 2). All values are estimated. f of negative 1 is unknown.
  10. 86.

    The graph of y = f of x is a line that rises from a closed point at (negative 3, negative 5) through (0, 1) to a closed point at (2, 5). All values are estimated. f of negative 2 is unknown.

  11. 87.

    The graph of y = f of x is a line that rises through (negative 4, negative 5) to (negative 1, negative 2).
  12. 88.

    The graph of y = f of x is a line that passes along the negative x axis till the origin and then rises through (1, 2) and (2, 4). All values are estimated. f of negative 2 and f of 2 are unknown.
  13. 89.

    The graph of y = f of x is concave up decreasing curve with indefinite ends.
  14. 90.

    The graph of y = f of x includes two curves.
  15. 91.

    The graph of y = f of x plots five points as follows, (negative 5, 2), (negative 2, 2), (0, 2), (1, 2), and (3, 2). All values are estimated. f of negative 5 + f of 3 is unknown.
  16. 92.

    The graph of y = f of x plots 5 points at (negative 5, negative 2), (negative 2, negative 2), (0, negative 2), (1, negative 2), and (4, negative 2). All values are estimated. f of negative 5 + f of 4 is unknown.

Practice PLUS

In Exercises 9394, let f(x)=x2x+4 and g(x)=3x5.

  1. 93. Find g(1) and f(g(1)).

  2. 94. Find g(1) and f(g(1)).

In Exercises 9596, let f and g be defined by the following table:

x f(x) g(x)
2 6 0
1 3 4
0 1 1
1 4 3
2 0 6
  1. 95. Find f(1)f(0)[g(2)]2+f(2)÷g(2)g(1).

  2. 96. Find |f(1)f(0)|[g(1)]2+g(1)÷f(1)g(2).

In Exercises 9798, find f(x)f(x) for the given function f. Then simplify the expression.

  1. 97. f(x)=x3+x5

  2. 98. f(x)=x23x+7

Application Exercises

The bar graph shows public spending by the top five and the bottom five countries on pre-primary education and child care. Spending is given by public expenditure as a percentage of gross domestic product. Use the graph to solve Exercises 99100.

A graph titled top five and bottom five countries spending on pre primary education and child care.

Source: USA Today

  1. 99.

    1. Write a set of five ordered pairs in which each of the top-spending countries corresponds to spending as a percentage of gross domestic product. Each ordered pair should be of the form

       (country, percentage of gross domestic product).


    2. Is the relation in part (a) a function? Explain your answer.

    3. For the five top-spending countries, write a set of ordered pairs in which its spending as a percentage of gross domestic product corresponds to the country. Each ordered pair should be of the form

       (percentage of gross domestic product, country).


    4. Is the relation in part (c) a function? Explain your answer.

  2. 100.

    1. Write a set of five ordered pairs in which each of the bottom-spending countries corresponds to spending as a percentage of gross domestic product. Each ordered pair should be of the form

       (country, percentage of gross domestic product).


    2. Is the relation in part (a) a function? Explain your answer.

    3. For the five bottom-spending countries, write a set of ordered pairs in which its spending as a percentage of gross domestic product corresponds to the country. Each ordered pair should be of the form

       (percentage of gross domestic product, country).


    4. Is the relation in part (c) a function? Explain your answer.

The bar graph shows your chances of surviving to various ages once you reach 60.

A graph titled chances of 60 year olds surviving to various ages.

Source: National Center for Health Statistics

The functions

f(x)=2.9x+286andg(x)=0.01x24.9x+370.

model the chance, as a percent, that a 60-year-old will survive to age x. Use this information to solve Exercises 101102.

  1. 101.

    1. Find and interpret f(70).

    2. Find and interpret g(70).

    3. Which function serves as a better model for the chance of surviving to age 70?

  2. 102.

    1. Find and interpret f(90).

    2. Find and interpret g(90).

    3. Which function serves as a better model for the chance of surviving to age 90?

The wage gap, which is used to compare the status of women’s earnings relative to men’s, is expressed as a percent and is calculated by dividing the median, or middlemost, annual earnings for all women by the median annual earnings for all men. The bar graph shows the wage gap for selected years from 1980 through 2020.

A graph titled median women's earnings as a percentage of median men's earnings in the United States.
A graph titled the graph of a function modeling the data, plots wage gap, in percent, versus year after 19 80.

Source: Bureau of Labor Statistics

The function G(x)=0.01x2+0.9x+63 models the wage gap, as a percent, x years after 1980. The graph of function G is shown to the right of the actual data. Use this information to solve Exercises 103104.

  1. 103.

    1. Find and interpret G(40). Identify this information as a point on the graph of the function.

    2. Does G(40) overestimate or underestimate the actual data shown by the bar graph? By how much?

  2. 104.

    1. Find and interpret G(10). Identify this information as a point on the graph of the function.

    2. Does G(10) overestimate or underestimate the actual data shown by the bar graph? By how much?

In Exercises 105108, you will be developing functions that model given conditions.

  1. 105. A company that manufactures bicycles has a fixed cost of $100,000. It costs $100 to produce each bicycle. The total cost for the company is the sum of its fixed cost and variable costs. Write the total cost, C, as a function of the number of bicycles produced, x. Then find and interpret C(90).

  2. 106. A previously owned car was purchased for $22,500. The value of the car decreased by $3200 per year for each of the next six years. Write a function that describes the value of the car, V, after x years, where 0x6. Then find and interpret V(3).

  3. 107. You commute to work a distance of 40 miles and return on the same route at the end of the day. Your average rate on the return trip is 30 miles per hour faster than your average rate on the outgoing trip. Write the total time, T, in hours, devoted to your outgoing and return trips as a function of your rate on the outgoing trip, x. Then find and interpret T(30). Hint:

    Time traveled=Distance traveledRate of travel.
  4. 108. A chemist working on a flu vaccine needs to mix a 10% sodium-iodine solution with a 60% sodium-iodine solution to obtain a 50-milliliter mixture. Write the amount of sodium iodine in the mixture, S, in milliliters, as a function of the number of milliliters of the 10% solution used, x. Then find and interpret S(30).

Explaining the Concepts

  1. 109. What is a relation? Describe what is meant by its domain and its range.

  2. 110. Explain how to determine whether a relation is a function. What is a function?

  3. 111. How do you determine if an equation in x and y defines y as a function of x?

  4. 112. Does f(x) mean f times x when referring to a function f? If not, what does f(x) mean? Provide an example with your explanation.

  5. 113. What is the graph of a function?

  6. 114. Explain how the vertical line test is used to determine whether a graph represents a function.

  7. 115. Explain how to identify the domain and range of a function from its graph.

  8. 116. For people filing a single return, federal income tax is a function of adjusted gross income because for each value of adjusted gross income there is a specific tax to be paid. By contrast, the price of a house is not a function of the lot size on which the house sits because houses on same-sized lots can sell for many different prices.

    1. Describe an everyday situation between variables that is a function.

    2. Describe an everyday situation between variables that is not a function.

Technology Exercise

  1. 117. Use a graphing utility to verify any five pairs of graphs that you drew by hand in Exercises 3954.

Critical Thinking Exercises

MAKE SENSE? In Exercises 118121, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 118. My body temperature is a function of the time of day.

  2. 119. Using f(x)=3x+2, I found f(50) by applying the distributive property to (3x+2)50.

  3. 120. I graphed a function showing how paid vacation days depend on the number of years a person works for a company. The domain was the number of paid vacation days.

  4. 121. I graphed a function showing how the average number of annual physician visits depends on a person’s age. The domain was the average number of annual physician visits.

Use the graph of f to determine whether each statement in Exercises 122125 is true or false.

The graph of f plots a curve and a line.
  1. 122. The domain of f is [4, 1)(1, 4].

  2. 123. The range of f is [2, 2].

  3. 124. f(1)f(4)=2

  4. 125. f(0)=2.1

  5. 126. If f(x)=3x+7, find f(a+h)f(a)h.

  6. 127. Give an example of a relation with the following characteristics: The relation is a function containing two ordered pairs. Reversing the components in each ordered pair results in a relation that is not a function.

  7. 128. If f(x+y)=f(x)+f(y) and f(1)=3, find f(2), f(3), and f(4). Is f(x+y)=f(x)+f(y) for all functions?

Retaining the Concepts

  1. 129. Solve and check: 1 + 3(x4) = 2x. (Section P.7, Example 1)

  2. 130. Solve and check: x35x42=5.

    (Section P.7, Example 2)

  3. 131. Sharks may be scary, but they were responsible for only three deaths worldwide in 2014. The world’s deadliest creatures, ranked by the number of human deaths per year, are mosquitoes, snails, and snakes. The number of deaths by mosquitoes exceeds the number of deaths by snakes by 661 thousand. The number of deaths by snails exceeds the number of deaths by snakes by 106 thousand. Combined, mosquitoes, snails, and snakes result in 1049 thousand (or 1,049,000) human deaths per year. Determine the number of human deaths per year, in thousands, by snakes, mosquitoes, and snails. (Source: World Health Organization) (Section P.8, Example 1)

Preview Exercises

Exercises 132134 will help you prepare for the material covered in the next section.

  1. 132. The function C(g)=40+0.60(g  15) describes the monthly cost, C(g), in dollars, for a high-speed wireless Internet plan for g gigabytes of data, where g>15. Find and interpret C(45).

  2. 133. Use point plotting to graph f(x)=x+2 if x1.

  3. 134. Simplify: 2(x+h)2+3(x+h)+5(2x2+3x+5).

Section 1.3: More on Functions and Their Graphs

Section 1.3 More on Functions and Their Graphs

Learning Objectives

What You’ll Learn

  1. 1 Identify intervals on which a function increases, decreases, or is constant.

  2. 2 Use graphs to locate relative maxima or minima.

  3. 3 Test for symmetry.

  4. 4 Identify even or odd functions and recognize their symmetries.

  5. 5 Understand and use piecewise functions.

  6. 6 Find and simplify a function’s difference quotient.

It’s hard to believe that this gas-guzzler, with its huge fins and overstated design, was available in 1957 for approximately $1800. The line graph in Figure 1.28 shows the average fuel efficiency, in miles per gallon, of new U.S. passenger cars for selected years from 1955 through 2018. Based on the averages shown by the graph, it’s unlikely that this classic 1957 Chevy got more than 15 miles per gallon.

Figure 1.28

A graph titled, the average fuel efficiency of new U.S. passenger cars.

Source: U.S. Department of Transportation

Figure 1.28 Full Alternative Text

You are probably familiar with the words used to describe the graph in Figure 1.28:

Decreased, increased, minimum, and maximum.

In this section, you will enhance your intuitive understanding of ways of describing graphs by viewing these descriptions from the perspective of functions.

Objective 1: Identify intervals on which a function increases, decreases, or is constant

Increasing and Decreasing Functions

  1. Objective 1 Identify intervals on which a function increases, decreases, or is constant.

Watch Video

Too late for that flu shot now! It’s only 8 a.m. and you’re feeling lousy. Your temperature is 101°F. Fascinated by the way that algebra models the world (your author is projecting a bit here), you decide to construct graphs showing your body temperature as a function of the time of day. You decide to let x represent the number of hours after 8 a.m. and f(x) your temperature at time x.

At 8 a.m. your temperature is 101°F and you are not feeling well. However, your temperature starts to decrease. It reaches normal (98.6°F) by 11 a.m. Feeling energized, you construct the graph shown on the right, indicating decreasing temperature for {x|0<x<3}, or on the interval (0, 3).

A graph depicts that temperature decreases on (0, 3), reaching 98.6 degrees by 11 A.M.
1.3-123 Full Alternative Text

Did creating that first graph drain you of your energy? Your temperature starts to rise after 11 a.m. By 1 p.m., 5 hours after 8 a.m., your temperature reaches 100°F. However, you keep plotting points on your graph. At the right, we can see that your temperature increases for {x|3<x<5}, or on the interval (3, 5).

A graph depicts that temperature increases on (3, 5).
1.3-124 Full Alternative Text

The graph of f is decreasing to the left of x=3 and increasing to the right of x=3. Thus, your temperature 3 hours after 8 a.m. was at its lowest point. Your relative minimum temperature was 98.6°.

By 3 p.m., your temperature is no worse than it was at 1 p.m.: It is still 100°F. (Of course, it’s no better either.) Your temperature remained the same, or constant, for {x|5<x<7}, or on the interval (5, 7).

A graph depicts that temperature remains constant at 100 degrees on (5, 7).
1.3-125 Full Alternative Text

The time-temperature flu scenario illustrates that a function f is increasing when its graph rises from left to right, decreasing when its graph falls from left to right, and remains constant when it neither rises nor falls. Let’s now provide a more precise algebraic description for these intuitive concepts.

Increasing, Decreasing, and Constant Functions

  1. A function is increasing on an open interval, I, if f(x1)<f(x2) whenever x1<x2 for any x1 and x2 in the interval.

  2. A function is decreasing on an open interval, I, if f(x1)>f(x2) whenever x1<x2 for any x1 and x2 in the interval.

  3. A function is constant on an open interval, I, if f(x1)=f(x2) for any x1 and x2 in the interval.

The graph plots an increasing curve that rises from left to right.
Graph 2 plots a decreasing curve that falls from left to right.
The graph plots a constant line that runs horizontally from left to right.

Example 1 Intervals on Which a Function Increases, Decreases, or Is Constant

State the intervals on which each given function is increasing, decreasing, or constant.

  1.  

    The graph of f of x = 3 x squared minus x cubed is a curve with indefinite ends that falls through (negative 1, 4) to (0, 0) and then rises through (1, 2) to (2, 4), after which it falls through (3, 0) and (3.5, negative 6). All values are estimated.
  2.  

    A graph of y = f of x plots a horizontal line and a curve.

Solution

  1. The function is decreasing on the interval (, 0), increasing on the interval (0, 2), and decreasing on the interval (2, ).

  2. Although the function’s equations are not given, the graph indicates that the function is defined in two pieces. The part of the graph to the left of the y-axis shows that the function is constant on the interval (, 0). The part to the right of the y-axis shows that the function is increasing on the interval (0, ).

Check Point 1

Objective 1: Identify intervals on which a function increases, decreases, or is constant

Increasing and Decreasing Functions

  1. Objective 1 Identify intervals on which a function increases, decreases, or is constant.

Watch Video

Too late for that flu shot now! It’s only 8 a.m. and you’re feeling lousy. Your temperature is 101°F. Fascinated by the way that algebra models the world (your author is projecting a bit here), you decide to construct graphs showing your body temperature as a function of the time of day. You decide to let x represent the number of hours after 8 a.m. and f(x) your temperature at time x.

At 8 a.m. your temperature is 101°F and you are not feeling well. However, your temperature starts to decrease. It reaches normal (98.6°F) by 11 a.m. Feeling energized, you construct the graph shown on the right, indicating decreasing temperature for {x|0<x<3}, or on the interval (0, 3).

A graph depicts that temperature decreases on (0, 3), reaching 98.6 degrees by 11 A.M.
1.3-123 Full Alternative Text

Did creating that first graph drain you of your energy? Your temperature starts to rise after 11 a.m. By 1 p.m., 5 hours after 8 a.m., your temperature reaches 100°F. However, you keep plotting points on your graph. At the right, we can see that your temperature increases for {x|3<x<5}, or on the interval (3, 5).

A graph depicts that temperature increases on (3, 5).
1.3-124 Full Alternative Text

The graph of f is decreasing to the left of x=3 and increasing to the right of x=3. Thus, your temperature 3 hours after 8 a.m. was at its lowest point. Your relative minimum temperature was 98.6°.

By 3 p.m., your temperature is no worse than it was at 1 p.m.: It is still 100°F. (Of course, it’s no better either.) Your temperature remained the same, or constant, for {x|5<x<7}, or on the interval (5, 7).

A graph depicts that temperature remains constant at 100 degrees on (5, 7).
1.3-125 Full Alternative Text

The time-temperature flu scenario illustrates that a function f is increasing when its graph rises from left to right, decreasing when its graph falls from left to right, and remains constant when it neither rises nor falls. Let’s now provide a more precise algebraic description for these intuitive concepts.

Increasing, Decreasing, and Constant Functions

  1. A function is increasing on an open interval, I, if f(x1)<f(x2) whenever x1<x2 for any x1 and x2 in the interval.

  2. A function is decreasing on an open interval, I, if f(x1)>f(x2) whenever x1<x2 for any x1 and x2 in the interval.

  3. A function is constant on an open interval, I, if f(x1)=f(x2) for any x1 and x2 in the interval.

The graph plots an increasing curve that rises from left to right.
Graph 2 plots a decreasing curve that falls from left to right.
The graph plots a constant line that runs horizontally from left to right.

Example 1 Intervals on Which a Function Increases, Decreases, or Is Constant

State the intervals on which each given function is increasing, decreasing, or constant.

  1.  

    The graph of f of x = 3 x squared minus x cubed is a curve with indefinite ends that falls through (negative 1, 4) to (0, 0) and then rises through (1, 2) to (2, 4), after which it falls through (3, 0) and (3.5, negative 6). All values are estimated.
  2.  

    A graph of y = f of x plots a horizontal line and a curve.

Solution

  1. The function is decreasing on the interval (, 0), increasing on the interval (0, 2), and decreasing on the interval (2, ).

  2. Although the function’s equations are not given, the graph indicates that the function is defined in two pieces. The part of the graph to the left of the y-axis shows that the function is constant on the interval (, 0). The part to the right of the y-axis shows that the function is increasing on the interval (0, ).

Check Point 1

Objective 2: Use graphs to locate relative maxima or minima

Relative Maxima and Relative Minima

  1. Objective 2 Use graphs to locate relative maxima or minima.

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The points at which a function changes its increasing or decreasing behavior can be used to find any relative maximum or relative minimum values of the function.

Definitions of Relative Maximum and Relative Minimum

  1. A function value f(a) is a relative maximum of f if there exists an open interval containing a such that f(a)>f(x) for all xa in the open interval.

  2. A function value f(b) is a relative minimum of f if there exists an open interval containing b such that f(b)<f(x) for all xb in the open interval.

A graph plots a curve.

The word local is sometimes used instead of relative when describing maxima or minima.

If the graph of a function is given, we can often visually locate the number(s) at which the function has a relative maximum or a relative minimum. For example, the graph of f in Figure 1.29 shows that

Figure 1.29Using a graph to locate where a function has a relative maximum or minimum

A graph of f plots a sinusoidal curve with indefinite ends of amplitude 1 and periodicity pi.

Figure 1.29 Full Alternative Text
Objective 3: Test for symmetry

Tests for Symmetry

  1. Objective 3 Test for symmetry.

Watch Video

Is beauty in the eye of the beholder? Or are there certain objects (or people) that are so well balanced and proportioned that they are universally pleasing to the eye? What constitutes an attractive human face? In Figure 1.30, we’ve drawn lines between paired features. Notice how the features line up almost perfectly. Each half of the face is a mirror image of the other half through the vertical line. If we superimpose a rectangular coordinate system on the attractive face, notice that a point (x, y) on the right has a mirror image at (x, y) on the left. The attractive face is said to be symmetric with respect to the y-axis.

Figure 1.30To most people, an attractive face is one in which each half is an almost perfect mirror image of the other half.

A rectangular coordinate system superimposes a man's face. His left pupil is at (x, y), and his right pupil is (negative x, y).
Figure 1.30 Full Alternative Text

Did you know that graphs of some equations exhibit exactly the kind of symmetry shown by the attractive face in Figure 1.30, as well as other kinds of symmetry? The word symmetry comes from the Greek symmetria, meaning “the same measure.” Figure 1.31 shows three graphs, each with a common form of symmetry. Notice that the graph in Figure 1.31(a) shows the y-axis symmetry found in the attractive face.

Figure 1.31Graphs illustrating the three forms of symmetry

The image shows three graphs.
Figure 1.31 Full Alternative Text

Table 1.2 defines three common forms of symmetry and gives rules to determine if the graph of an equation is symmetric with respect to the y-axis, the x-axis, or the origin.

Table 1.2 Definitions and Tests for Symmetry

A table lists the definitions of and the tests for symmetry.

Consider an equation in the variables x and y.

Table 1.2 Full Alternative Text

Example 2 Testing for Symmetry

Determine whether the graph of

x=y21

is symmetric with respect to the y-axis, the x-axis, or the origin.

Solution

Test for symmetry with respect to the y-axis. Replace x with x and see if this results in an equivalent equation.

The given equation. x = y squared minus 1. Replace x with negative x. Negative x = y squared minus 1. Multiply both sides by negative 1 and solve for x. x = negative y squared + 1. The first and third equations are not equivalent.

Replacing x with x does not give the original equation. Thus, the graph of x=y21 is not symmetric with respect to the y-axis.

Test for symmetry with respect to the x-axis. Replace y with y and see if this results in an equivalent equation.

The process to get an equivalent equation for x = y squared minus 1.

Replacing y with y gives the original equation. Thus, the graph of x=y21 is symmetric with respect to the x-axis.

Test for symmetry with respect to the origin. Replace x with x and y with y and see if this results in an equivalent equation.

The process to get an equation that is not equivalent to x = y squared minus 1.

Replacing x with x and y with y does not give the original equation. Thus, the graph of x=y21 is not symmetric with respect to the origin.

Our symmetry tests reveal that the graph of x=y21 is symmetric with respect to the x-axis only.

The graph of x=y21 in Figure 1.32(a) shows the symmetry with respect to the x-axis that we determined in Example 2. Figure 1.32(b) illustrates that the graph fails the vertical line test. Consequently, y is not a function of x. We have seen that if a graph has x-axis symmetry, y is usually not a function of x.

Figure 1.32 The graph of x=y21

Two graphs, a and b, plot x = y squared minus 1, which is a rightward opening parabola that falls through (3, 2) with its vertex at (negative 1, 1) and then continues to fall through (3, negative 2).

Check Point 2

  • Determine whether the graph of y=x21 is symmetric with respect to the y-axis, the x-axis, or the origin.

Example 3 Testing for Symmetry

Determine whether the graph of y=x3 is symmetric with respect to the y-axis, the x-axis, or the origin.

Solution

Test for symmetry with respect to the y-axis. Replace x with x and see if this results in an equivalent equation.

The process to get an equation that is not equivalent to y = x cubed.

Replacing x with x does not give the original equation. Thus, the graph of y=x3 is not symmetric with respect to the y-axis.

Test for symmetry with respect to the x-axis. Replace y with y and see if this results in an equivalent equation.

The given equation. y = x cubed. Replace y with negative y. Negative y = x cubed. Multiply both sides by negative 1 and solve for y. y = negative x cubed. The first and third equations are not the same.

Replacing y with y does not give the original equation. Thus, the graph of y=x3 is not symmetric with respect to the x-axis.

Test for symmetry with respect to the origin. Replace x with x and y with y and see if this results in an equivalent equation.

The image shows a mathematical expression to find the value of y.

Replacing x with x and y with y gives the original equation. Thus, the graph of y=x3 is symmetric with respect to the origin.

Our symmetry tests reveal that the graph of y=x3 is symmetric with respect to the origin only. The symmetry is shown in Figure 1.33.

Figure 1.33 The graph is symmetric with respect to the origin.

An image shows that the graph is symmetric with respect to the origin.

Check Point 3

  • Determine whether the graph of y5=x3 is symmetric with respect to the y-axis, the x-axis, or the origin.

Objective 3: Test for symmetry

Tests for Symmetry

  1. Objective 3 Test for symmetry.

Watch Video

Is beauty in the eye of the beholder? Or are there certain objects (or people) that are so well balanced and proportioned that they are universally pleasing to the eye? What constitutes an attractive human face? In Figure 1.30, we’ve drawn lines between paired features. Notice how the features line up almost perfectly. Each half of the face is a mirror image of the other half through the vertical line. If we superimpose a rectangular coordinate system on the attractive face, notice that a point (x, y) on the right has a mirror image at (x, y) on the left. The attractive face is said to be symmetric with respect to the y-axis.

Figure 1.30To most people, an attractive face is one in which each half is an almost perfect mirror image of the other half.

A rectangular coordinate system superimposes a man's face. His left pupil is at (x, y), and his right pupil is (negative x, y).
Figure 1.30 Full Alternative Text

Did you know that graphs of some equations exhibit exactly the kind of symmetry shown by the attractive face in Figure 1.30, as well as other kinds of symmetry? The word symmetry comes from the Greek symmetria, meaning “the same measure.” Figure 1.31 shows three graphs, each with a common form of symmetry. Notice that the graph in Figure 1.31(a) shows the y-axis symmetry found in the attractive face.

Figure 1.31Graphs illustrating the three forms of symmetry

The image shows three graphs.
Figure 1.31 Full Alternative Text

Table 1.2 defines three common forms of symmetry and gives rules to determine if the graph of an equation is symmetric with respect to the y-axis, the x-axis, or the origin.

Table 1.2 Definitions and Tests for Symmetry

A table lists the definitions of and the tests for symmetry.

Consider an equation in the variables x and y.

Table 1.2 Full Alternative Text

Example 2 Testing for Symmetry

Determine whether the graph of

x=y21

is symmetric with respect to the y-axis, the x-axis, or the origin.

Solution

Test for symmetry with respect to the y-axis. Replace x with x and see if this results in an equivalent equation.

The given equation. x = y squared minus 1. Replace x with negative x. Negative x = y squared minus 1. Multiply both sides by negative 1 and solve for x. x = negative y squared + 1. The first and third equations are not equivalent.

Replacing x with x does not give the original equation. Thus, the graph of x=y21 is not symmetric with respect to the y-axis.

Test for symmetry with respect to the x-axis. Replace y with y and see if this results in an equivalent equation.

The process to get an equivalent equation for x = y squared minus 1.

Replacing y with y gives the original equation. Thus, the graph of x=y21 is symmetric with respect to the x-axis.

Test for symmetry with respect to the origin. Replace x with x and y with y and see if this results in an equivalent equation.

The process to get an equation that is not equivalent to x = y squared minus 1.

Replacing x with x and y with y does not give the original equation. Thus, the graph of x=y21 is not symmetric with respect to the origin.

Our symmetry tests reveal that the graph of x=y21 is symmetric with respect to the x-axis only.

The graph of x=y21 in Figure 1.32(a) shows the symmetry with respect to the x-axis that we determined in Example 2. Figure 1.32(b) illustrates that the graph fails the vertical line test. Consequently, y is not a function of x. We have seen that if a graph has x-axis symmetry, y is usually not a function of x.

Figure 1.32 The graph of x=y21

Two graphs, a and b, plot x = y squared minus 1, which is a rightward opening parabola that falls through (3, 2) with its vertex at (negative 1, 1) and then continues to fall through (3, negative 2).

Check Point 2

  • Determine whether the graph of y=x21 is symmetric with respect to the y-axis, the x-axis, or the origin.

Example 3 Testing for Symmetry

Determine whether the graph of y=x3 is symmetric with respect to the y-axis, the x-axis, or the origin.

Solution

Test for symmetry with respect to the y-axis. Replace x with x and see if this results in an equivalent equation.

The process to get an equation that is not equivalent to y = x cubed.

Replacing x with x does not give the original equation. Thus, the graph of y=x3 is not symmetric with respect to the y-axis.

Test for symmetry with respect to the x-axis. Replace y with y and see if this results in an equivalent equation.

The given equation. y = x cubed. Replace y with negative y. Negative y = x cubed. Multiply both sides by negative 1 and solve for y. y = negative x cubed. The first and third equations are not the same.

Replacing y with y does not give the original equation. Thus, the graph of y=x3 is not symmetric with respect to the x-axis.

Test for symmetry with respect to the origin. Replace x with x and y with y and see if this results in an equivalent equation.

The image shows a mathematical expression to find the value of y.

Replacing x with x and y with y gives the original equation. Thus, the graph of y=x3 is symmetric with respect to the origin.

Our symmetry tests reveal that the graph of y=x3 is symmetric with respect to the origin only. The symmetry is shown in Figure 1.33.

Figure 1.33 The graph is symmetric with respect to the origin.

An image shows that the graph is symmetric with respect to the origin.

Check Point 3

  • Determine whether the graph of y5=x3 is symmetric with respect to the y-axis, the x-axis, or the origin.

Objective 3: Test for symmetry

Tests for Symmetry

  1. Objective 3 Test for symmetry.

Watch Video

Is beauty in the eye of the beholder? Or are there certain objects (or people) that are so well balanced and proportioned that they are universally pleasing to the eye? What constitutes an attractive human face? In Figure 1.30, we’ve drawn lines between paired features. Notice how the features line up almost perfectly. Each half of the face is a mirror image of the other half through the vertical line. If we superimpose a rectangular coordinate system on the attractive face, notice that a point (x, y) on the right has a mirror image at (x, y) on the left. The attractive face is said to be symmetric with respect to the y-axis.

Figure 1.30To most people, an attractive face is one in which each half is an almost perfect mirror image of the other half.

A rectangular coordinate system superimposes a man's face. His left pupil is at (x, y), and his right pupil is (negative x, y).
Figure 1.30 Full Alternative Text

Did you know that graphs of some equations exhibit exactly the kind of symmetry shown by the attractive face in Figure 1.30, as well as other kinds of symmetry? The word symmetry comes from the Greek symmetria, meaning “the same measure.” Figure 1.31 shows three graphs, each with a common form of symmetry. Notice that the graph in Figure 1.31(a) shows the y-axis symmetry found in the attractive face.

Figure 1.31Graphs illustrating the three forms of symmetry

The image shows three graphs.
Figure 1.31 Full Alternative Text

Table 1.2 defines three common forms of symmetry and gives rules to determine if the graph of an equation is symmetric with respect to the y-axis, the x-axis, or the origin.

Table 1.2 Definitions and Tests for Symmetry

A table lists the definitions of and the tests for symmetry.

Consider an equation in the variables x and y.

Table 1.2 Full Alternative Text

Example 2 Testing for Symmetry

Determine whether the graph of

x=y21

is symmetric with respect to the y-axis, the x-axis, or the origin.

Solution

Test for symmetry with respect to the y-axis. Replace x with x and see if this results in an equivalent equation.

The given equation. x = y squared minus 1. Replace x with negative x. Negative x = y squared minus 1. Multiply both sides by negative 1 and solve for x. x = negative y squared + 1. The first and third equations are not equivalent.

Replacing x with x does not give the original equation. Thus, the graph of x=y21 is not symmetric with respect to the y-axis.

Test for symmetry with respect to the x-axis. Replace y with y and see if this results in an equivalent equation.

The process to get an equivalent equation for x = y squared minus 1.

Replacing y with y gives the original equation. Thus, the graph of x=y21 is symmetric with respect to the x-axis.

Test for symmetry with respect to the origin. Replace x with x and y with y and see if this results in an equivalent equation.

The process to get an equation that is not equivalent to x = y squared minus 1.

Replacing x with x and y with y does not give the original equation. Thus, the graph of x=y21 is not symmetric with respect to the origin.

Our symmetry tests reveal that the graph of x=y21 is symmetric with respect to the x-axis only.

The graph of x=y21 in Figure 1.32(a) shows the symmetry with respect to the x-axis that we determined in Example 2. Figure 1.32(b) illustrates that the graph fails the vertical line test. Consequently, y is not a function of x. We have seen that if a graph has x-axis symmetry, y is usually not a function of x.

Figure 1.32 The graph of x=y21

Two graphs, a and b, plot x = y squared minus 1, which is a rightward opening parabola that falls through (3, 2) with its vertex at (negative 1, 1) and then continues to fall through (3, negative 2).

Check Point 2

  • Determine whether the graph of y=x21 is symmetric with respect to the y-axis, the x-axis, or the origin.

Example 3 Testing for Symmetry

Determine whether the graph of y=x3 is symmetric with respect to the y-axis, the x-axis, or the origin.

Solution

Test for symmetry with respect to the y-axis. Replace x with x and see if this results in an equivalent equation.

The process to get an equation that is not equivalent to y = x cubed.

Replacing x with x does not give the original equation. Thus, the graph of y=x3 is not symmetric with respect to the y-axis.

Test for symmetry with respect to the x-axis. Replace y with y and see if this results in an equivalent equation.

The given equation. y = x cubed. Replace y with negative y. Negative y = x cubed. Multiply both sides by negative 1 and solve for y. y = negative x cubed. The first and third equations are not the same.

Replacing y with y does not give the original equation. Thus, the graph of y=x3 is not symmetric with respect to the x-axis.

Test for symmetry with respect to the origin. Replace x with x and y with y and see if this results in an equivalent equation.

The image shows a mathematical expression to find the value of y.

Replacing x with x and y with y gives the original equation. Thus, the graph of y=x3 is symmetric with respect to the origin.

Our symmetry tests reveal that the graph of y=x3 is symmetric with respect to the origin only. The symmetry is shown in Figure 1.33.

Figure 1.33 The graph is symmetric with respect to the origin.

An image shows that the graph is symmetric with respect to the origin.

Check Point 3

  • Determine whether the graph of y5=x3 is symmetric with respect to the y-axis, the x-axis, or the origin.

Objective 4: Identify even or odd functions and recognize their symmetries

Even and Odd Functions

  1. Objective 4 Identify even or odd functions and recognize their symmetries.

Watch Video

We have seen that if a graph is symmetric with respect to the x-axis, it usually fails the vertical line test and is not the graph of a function. However, many functions have graphs that are symmetric with respect to the y-axis or the origin. We give these functions special names.

A function whose graph is symmetric with respect to the y-axis is called an even function. If the point (x, y) is on the graph of the function, then the point (x, y) is also on the graph. Expressing this symmetry in function notation provides the definition of an even function.

Even Functions and Their Symmetries

The function f is an even function if

f(x)=f(x)for all x in the domain of f.

The graph of an even function is symmetric with respect to the y-axis.

An example of an even function is f(x)=x2. The graph, shown in Figure 1.35, is symmetric with respect to the y-axis.

Figure 1.35 The graph of f(x)=x2, an even function

The graph of f of x = x squared is an upward opening parabola that falls through (negative 3, 9) and (negative x, f of x) to its vertex at (0, 0) and rises through (x, f of x) and (3, 9).

Figure 1.35 Full Alternative Text

Let’s show that f(x)=f(x).

f(x)=x2This is the functions equation.f(x)=(x)2Replace x with x.f(x)=x2(x)2=(x)(x)=x2

Because f(x)=x2 and  f(x)=x2, we see that f(x)=f(x). This algebraically identifies the function as an even function.

A function whose graph is symmetric with respect to the origin is called an odd function. If the point (x, y) is on the graph of the function, then the point (x, y) is also on the graph. Expressing this symmetry in function notation provides the definition of an odd function.

Odd Functions and Their Symmetries

The function f is an odd function if

f(x)=f(x)for all x in the domain of f.

The graph of an odd function is symmetric with respect to the origin.

An example of an odd function is f(x)=x3. The graph, shown in Figure 1.36, is symmetric with respect to the origin.

Figure 1.36The graph of f(x)=x3, an odd function

The graph of f of x = x cubed is a curve that rises through (negative 2, negative 8), (negative x, negative f of x), (negative 0.5, 0), (0, 0), (0.5, 0), (x, f of x) and (2, 8).
Figure 1.36 Full Alternative Text

Let’s show that f(x)=f(x).

f(x)=x3This is the functions equation.f(x)=(x)3Replace x with x.f(x)=x3(x)3=(x)(x)(x)=x3

Because f(x)=x3 and f(x)=x3, we see that f(x)=f(x). This algebraically identifies the function as an odd function.

Example 4 Identifying Even or Odd Functions from Graphs

Determine whether each graph is the graph of an even function, an odd function, or a function that is neither even nor odd.

The image shows three graphs, a, b, and c.

Solution

Note that each graph passes the vertical line test and is therefore the graph of a function. In each case, use inspection to determine whether or not there is symmetry. If the graph is symmetric with respect to the y-axis, it is that of an even function. If the graph is symmetric with respect to the origin, it is that of an odd function.

The image shows three graphs, a, b, and c.

Check Point 4

  • Determine whether each graph is the graph of an even function, an odd function, or a function that is neither even nor odd.

    The image shows three graphs, a, b, and c.

In addition to inspecting graphs, we can also use a function’s equation and the definitions of even and odd to determine whether the function is even, odd, or neither.

Identifying Even or Odd Functions from Equations

Example 5 Identifying Even or Odd Functions from Equations

Determine whether each of the following functions is even, odd, or neither. Then determine whether the function’s graph is symmetric with respect to the y-axis, the origin, or neither.

  1. f(x)=x36x

  2. g(x)=x42x2

  3. h(x)=x2+2x+1

Solution

In each case, replace x with x and simplify. If the right side of the equation stays the same, the function is even. If every term on the right side changes sign, the function is odd.

  1. We use the given function’s equation, f(x)=x36x, to find f(x).

    Use f of x = x cubed minus x.

    There are two terms on the right side of the given equation, f(x)=x36x, and each term changed sign when we replaced x with x. Because f(x)=f(x), f is an odd function. The graph of f is symmetric with respect to the origin.

  2. We use the given function’s equation, g(x)=x42x2, to find g(x).

    The image shows a mathematical expression to simplify g of x = x to the fourth power minus 2 x squared.

    The right side of the equation of the given function, g(x)=x42x2, did not change when we replaced x with x. Because g(x)=g(x), g is an even function. The graph of g is symmetric with respect to the y-axis.

  3. We use the given function’s equation, h(x)=x2+2x+1, to find h(x).

    Use h of x = x squared + 2 x + 1. Replace x with negative x. h of negative x = left parenthesis negative x right parenthesis squared + 2 left parenthesis negative x right parenthesis + 1 = x squared minus 2 x + 1.

    The right side of the equation of the given function, h(x)=x2+2x+1, changed when we replaced x with x. Thus, h(x)h(x), so h is not an even function. The sign of each of the three terms in the equation for h(x) did not change when we replaced x with x. Only the second term changed signs. Thus, h(x)h(x), so h is not an odd function. We conclude that h is neither an even nor an odd function. The graph of h is neither symmetric with respect to the y-axis nor with respect to the origin.

Check Point 5

  • Determine whether each of the following functions is even, odd, or neither. Then determine whether the function’s graph is symmetric with respect to the y-axis, the origin, or neither.

    1. f(x)=x2+6

    2. g(x)=7x3x

    3. h(x)=x5+1

Objective 4: Identify even or odd functions and recognize their symmetries

Even and Odd Functions

  1. Objective 4 Identify even or odd functions and recognize their symmetries.

Watch Video

We have seen that if a graph is symmetric with respect to the x-axis, it usually fails the vertical line test and is not the graph of a function. However, many functions have graphs that are symmetric with respect to the y-axis or the origin. We give these functions special names.

A function whose graph is symmetric with respect to the y-axis is called an even function. If the point (x, y) is on the graph of the function, then the point (x, y) is also on the graph. Expressing this symmetry in function notation provides the definition of an even function.

Even Functions and Their Symmetries

The function f is an even function if

f(x)=f(x)for all x in the domain of f.

The graph of an even function is symmetric with respect to the y-axis.

An example of an even function is f(x)=x2. The graph, shown in Figure 1.35, is symmetric with respect to the y-axis.

Figure 1.35 The graph of f(x)=x2, an even function

The graph of f of x = x squared is an upward opening parabola that falls through (negative 3, 9) and (negative x, f of x) to its vertex at (0, 0) and rises through (x, f of x) and (3, 9).

Figure 1.35 Full Alternative Text

Let’s show that f(x)=f(x).

f(x)=x2This is the functions equation.f(x)=(x)2Replace x with x.f(x)=x2(x)2=(x)(x)=x2

Because f(x)=x2 and  f(x)=x2, we see that f(x)=f(x). This algebraically identifies the function as an even function.

A function whose graph is symmetric with respect to the origin is called an odd function. If the point (x, y) is on the graph of the function, then the point (x, y) is also on the graph. Expressing this symmetry in function notation provides the definition of an odd function.

Odd Functions and Their Symmetries

The function f is an odd function if

f(x)=f(x)for all x in the domain of f.

The graph of an odd function is symmetric with respect to the origin.

An example of an odd function is f(x)=x3. The graph, shown in Figure 1.36, is symmetric with respect to the origin.

Figure 1.36The graph of f(x)=x3, an odd function

The graph of f of x = x cubed is a curve that rises through (negative 2, negative 8), (negative x, negative f of x), (negative 0.5, 0), (0, 0), (0.5, 0), (x, f of x) and (2, 8).
Figure 1.36 Full Alternative Text

Let’s show that f(x)=f(x).

f(x)=x3This is the functions equation.f(x)=(x)3Replace x with x.f(x)=x3(x)3=(x)(x)(x)=x3

Because f(x)=x3 and f(x)=x3, we see that f(x)=f(x). This algebraically identifies the function as an odd function.

Example 4 Identifying Even or Odd Functions from Graphs

Determine whether each graph is the graph of an even function, an odd function, or a function that is neither even nor odd.

The image shows three graphs, a, b, and c.

Solution

Note that each graph passes the vertical line test and is therefore the graph of a function. In each case, use inspection to determine whether or not there is symmetry. If the graph is symmetric with respect to the y-axis, it is that of an even function. If the graph is symmetric with respect to the origin, it is that of an odd function.

The image shows three graphs, a, b, and c.

Check Point 4

  • Determine whether each graph is the graph of an even function, an odd function, or a function that is neither even nor odd.

    The image shows three graphs, a, b, and c.

In addition to inspecting graphs, we can also use a function’s equation and the definitions of even and odd to determine whether the function is even, odd, or neither.

Identifying Even or Odd Functions from Equations

Example 5 Identifying Even or Odd Functions from Equations

Determine whether each of the following functions is even, odd, or neither. Then determine whether the function’s graph is symmetric with respect to the y-axis, the origin, or neither.

  1. f(x)=x36x

  2. g(x)=x42x2

  3. h(x)=x2+2x+1

Solution

In each case, replace x with x and simplify. If the right side of the equation stays the same, the function is even. If every term on the right side changes sign, the function is odd.

  1. We use the given function’s equation, f(x)=x36x, to find f(x).

    Use f of x = x cubed minus x.

    There are two terms on the right side of the given equation, f(x)=x36x, and each term changed sign when we replaced x with x. Because f(x)=f(x), f is an odd function. The graph of f is symmetric with respect to the origin.

  2. We use the given function’s equation, g(x)=x42x2, to find g(x).

    The image shows a mathematical expression to simplify g of x = x to the fourth power minus 2 x squared.

    The right side of the equation of the given function, g(x)=x42x2, did not change when we replaced x with x. Because g(x)=g(x), g is an even function. The graph of g is symmetric with respect to the y-axis.

  3. We use the given function’s equation, h(x)=x2+2x+1, to find h(x).

    Use h of x = x squared + 2 x + 1. Replace x with negative x. h of negative x = left parenthesis negative x right parenthesis squared + 2 left parenthesis negative x right parenthesis + 1 = x squared minus 2 x + 1.

    The right side of the equation of the given function, h(x)=x2+2x+1, changed when we replaced x with x. Thus, h(x)h(x), so h is not an even function. The sign of each of the three terms in the equation for h(x) did not change when we replaced x with x. Only the second term changed signs. Thus, h(x)h(x), so h is not an odd function. We conclude that h is neither an even nor an odd function. The graph of h is neither symmetric with respect to the y-axis nor with respect to the origin.

Check Point 5

  • Determine whether each of the following functions is even, odd, or neither. Then determine whether the function’s graph is symmetric with respect to the y-axis, the origin, or neither.

    1. f(x)=x2+6

    2. g(x)=7x3x

    3. h(x)=x5+1

Objective 4: Identify even or odd functions and recognize their symmetries

Even and Odd Functions

  1. Objective 4 Identify even or odd functions and recognize their symmetries.

Watch Video

We have seen that if a graph is symmetric with respect to the x-axis, it usually fails the vertical line test and is not the graph of a function. However, many functions have graphs that are symmetric with respect to the y-axis or the origin. We give these functions special names.

A function whose graph is symmetric with respect to the y-axis is called an even function. If the point (x, y) is on the graph of the function, then the point (x, y) is also on the graph. Expressing this symmetry in function notation provides the definition of an even function.

Even Functions and Their Symmetries

The function f is an even function if

f(x)=f(x)for all x in the domain of f.

The graph of an even function is symmetric with respect to the y-axis.

An example of an even function is f(x)=x2. The graph, shown in Figure 1.35, is symmetric with respect to the y-axis.

Figure 1.35 The graph of f(x)=x2, an even function

The graph of f of x = x squared is an upward opening parabola that falls through (negative 3, 9) and (negative x, f of x) to its vertex at (0, 0) and rises through (x, f of x) and (3, 9).

Figure 1.35 Full Alternative Text

Let’s show that f(x)=f(x).

f(x)=x2This is the functions equation.f(x)=(x)2Replace x with x.f(x)=x2(x)2=(x)(x)=x2

Because f(x)=x2 and  f(x)=x2, we see that f(x)=f(x). This algebraically identifies the function as an even function.

A function whose graph is symmetric with respect to the origin is called an odd function. If the point (x, y) is on the graph of the function, then the point (x, y) is also on the graph. Expressing this symmetry in function notation provides the definition of an odd function.

Odd Functions and Their Symmetries

The function f is an odd function if

f(x)=f(x)for all x in the domain of f.

The graph of an odd function is symmetric with respect to the origin.

An example of an odd function is f(x)=x3. The graph, shown in Figure 1.36, is symmetric with respect to the origin.

Figure 1.36The graph of f(x)=x3, an odd function

The graph of f of x = x cubed is a curve that rises through (negative 2, negative 8), (negative x, negative f of x), (negative 0.5, 0), (0, 0), (0.5, 0), (x, f of x) and (2, 8).
Figure 1.36 Full Alternative Text

Let’s show that f(x)=f(x).

f(x)=x3This is the functions equation.f(x)=(x)3Replace x with x.f(x)=x3(x)3=(x)(x)(x)=x3

Because f(x)=x3 and f(x)=x3, we see that f(x)=f(x). This algebraically identifies the function as an odd function.

Example 4 Identifying Even or Odd Functions from Graphs

Determine whether each graph is the graph of an even function, an odd function, or a function that is neither even nor odd.

The image shows three graphs, a, b, and c.

Solution

Note that each graph passes the vertical line test and is therefore the graph of a function. In each case, use inspection to determine whether or not there is symmetry. If the graph is symmetric with respect to the y-axis, it is that of an even function. If the graph is symmetric with respect to the origin, it is that of an odd function.

The image shows three graphs, a, b, and c.

Check Point 4

  • Determine whether each graph is the graph of an even function, an odd function, or a function that is neither even nor odd.

    The image shows three graphs, a, b, and c.

In addition to inspecting graphs, we can also use a function’s equation and the definitions of even and odd to determine whether the function is even, odd, or neither.

Identifying Even or Odd Functions from Equations

Example 5 Identifying Even or Odd Functions from Equations

Determine whether each of the following functions is even, odd, or neither. Then determine whether the function’s graph is symmetric with respect to the y-axis, the origin, or neither.

  1. f(x)=x36x

  2. g(x)=x42x2

  3. h(x)=x2+2x+1

Solution

In each case, replace x with x and simplify. If the right side of the equation stays the same, the function is even. If every term on the right side changes sign, the function is odd.

  1. We use the given function’s equation, f(x)=x36x, to find f(x).

    Use f of x = x cubed minus x.

    There are two terms on the right side of the given equation, f(x)=x36x, and each term changed sign when we replaced x with x. Because f(x)=f(x), f is an odd function. The graph of f is symmetric with respect to the origin.

  2. We use the given function’s equation, g(x)=x42x2, to find g(x).

    The image shows a mathematical expression to simplify g of x = x to the fourth power minus 2 x squared.

    The right side of the equation of the given function, g(x)=x42x2, did not change when we replaced x with x. Because g(x)=g(x), g is an even function. The graph of g is symmetric with respect to the y-axis.

  3. We use the given function’s equation, h(x)=x2+2x+1, to find h(x).

    Use h of x = x squared + 2 x + 1. Replace x with negative x. h of negative x = left parenthesis negative x right parenthesis squared + 2 left parenthesis negative x right parenthesis + 1 = x squared minus 2 x + 1.

    The right side of the equation of the given function, h(x)=x2+2x+1, changed when we replaced x with x. Thus, h(x)h(x), so h is not an even function. The sign of each of the three terms in the equation for h(x) did not change when we replaced x with x. Only the second term changed signs. Thus, h(x)h(x), so h is not an odd function. We conclude that h is neither an even nor an odd function. The graph of h is neither symmetric with respect to the y-axis nor with respect to the origin.

Check Point 5

  • Determine whether each of the following functions is even, odd, or neither. Then determine whether the function’s graph is symmetric with respect to the y-axis, the origin, or neither.

    1. f(x)=x2+6

    2. g(x)=7x3x

    3. h(x)=x5+1

Objective 4: Identify even or odd functions and recognize their symmetries

Even and Odd Functions

  1. Objective 4 Identify even or odd functions and recognize their symmetries.

Watch Video

We have seen that if a graph is symmetric with respect to the x-axis, it usually fails the vertical line test and is not the graph of a function. However, many functions have graphs that are symmetric with respect to the y-axis or the origin. We give these functions special names.

A function whose graph is symmetric with respect to the y-axis is called an even function. If the point (x, y) is on the graph of the function, then the point (x, y) is also on the graph. Expressing this symmetry in function notation provides the definition of an even function.

Even Functions and Their Symmetries

The function f is an even function if

f(x)=f(x)for all x in the domain of f.

The graph of an even function is symmetric with respect to the y-axis.

An example of an even function is f(x)=x2. The graph, shown in Figure 1.35, is symmetric with respect to the y-axis.

Figure 1.35 The graph of f(x)=x2, an even function

The graph of f of x = x squared is an upward opening parabola that falls through (negative 3, 9) and (negative x, f of x) to its vertex at (0, 0) and rises through (x, f of x) and (3, 9).

Figure 1.35 Full Alternative Text

Let’s show that f(x)=f(x).

f(x)=x2This is the functions equation.f(x)=(x)2Replace x with x.f(x)=x2(x)2=(x)(x)=x2

Because f(x)=x2 and  f(x)=x2, we see that f(x)=f(x). This algebraically identifies the function as an even function.

A function whose graph is symmetric with respect to the origin is called an odd function. If the point (x, y) is on the graph of the function, then the point (x, y) is also on the graph. Expressing this symmetry in function notation provides the definition of an odd function.

Odd Functions and Their Symmetries

The function f is an odd function if

f(x)=f(x)for all x in the domain of f.

The graph of an odd function is symmetric with respect to the origin.

An example of an odd function is f(x)=x3. The graph, shown in Figure 1.36, is symmetric with respect to the origin.

Figure 1.36The graph of f(x)=x3, an odd function

The graph of f of x = x cubed is a curve that rises through (negative 2, negative 8), (negative x, negative f of x), (negative 0.5, 0), (0, 0), (0.5, 0), (x, f of x) and (2, 8).
Figure 1.36 Full Alternative Text

Let’s show that f(x)=f(x).

f(x)=x3This is the functions equation.f(x)=(x)3Replace x with x.f(x)=x3(x)3=(x)(x)(x)=x3

Because f(x)=x3 and f(x)=x3, we see that f(x)=f(x). This algebraically identifies the function as an odd function.

Example 4 Identifying Even or Odd Functions from Graphs

Determine whether each graph is the graph of an even function, an odd function, or a function that is neither even nor odd.

The image shows three graphs, a, b, and c.

Solution

Note that each graph passes the vertical line test and is therefore the graph of a function. In each case, use inspection to determine whether or not there is symmetry. If the graph is symmetric with respect to the y-axis, it is that of an even function. If the graph is symmetric with respect to the origin, it is that of an odd function.

The image shows three graphs, a, b, and c.

Check Point 4

  • Determine whether each graph is the graph of an even function, an odd function, or a function that is neither even nor odd.

    The image shows three graphs, a, b, and c.

In addition to inspecting graphs, we can also use a function’s equation and the definitions of even and odd to determine whether the function is even, odd, or neither.

Identifying Even or Odd Functions from Equations

Example 5 Identifying Even or Odd Functions from Equations

Determine whether each of the following functions is even, odd, or neither. Then determine whether the function’s graph is symmetric with respect to the y-axis, the origin, or neither.

  1. f(x)=x36x

  2. g(x)=x42x2

  3. h(x)=x2+2x+1

Solution

In each case, replace x with x and simplify. If the right side of the equation stays the same, the function is even. If every term on the right side changes sign, the function is odd.

  1. We use the given function’s equation, f(x)=x36x, to find f(x).

    Use f of x = x cubed minus x.

    There are two terms on the right side of the given equation, f(x)=x36x, and each term changed sign when we replaced x with x. Because f(x)=f(x), f is an odd function. The graph of f is symmetric with respect to the origin.

  2. We use the given function’s equation, g(x)=x42x2, to find g(x).

    The image shows a mathematical expression to simplify g of x = x to the fourth power minus 2 x squared.

    The right side of the equation of the given function, g(x)=x42x2, did not change when we replaced x with x. Because g(x)=g(x), g is an even function. The graph of g is symmetric with respect to the y-axis.

  3. We use the given function’s equation, h(x)=x2+2x+1, to find h(x).

    Use h of x = x squared + 2 x + 1. Replace x with negative x. h of negative x = left parenthesis negative x right parenthesis squared + 2 left parenthesis negative x right parenthesis + 1 = x squared minus 2 x + 1.

    The right side of the equation of the given function, h(x)=x2+2x+1, changed when we replaced x with x. Thus, h(x)h(x), so h is not an even function. The sign of each of the three terms in the equation for h(x) did not change when we replaced x with x. Only the second term changed signs. Thus, h(x)h(x), so h is not an odd function. We conclude that h is neither an even nor an odd function. The graph of h is neither symmetric with respect to the y-axis nor with respect to the origin.

Check Point 5

  • Determine whether each of the following functions is even, odd, or neither. Then determine whether the function’s graph is symmetric with respect to the y-axis, the origin, or neither.

    1. f(x)=x2+6

    2. g(x)=7x3x

    3. h(x)=x5+1

Objective 5: Understand and use piecewise functions

Piecewise Functions

  1. Objective 5 Understand and use piecewise functions.

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A cellphone company offers the following plan for a mobile hotspot:

We can represent this plan mathematically by writing the total monthly cost, C, as a function of the amount of data used, g.

C of g = 40, if g is greater than or equal to 0 and is less than or equal to 15 and 40 + 0.60 left parenthesis g minus 15 right parenthesis, if g is greater than 15.
1.3-148 Full Alternative Text

A function that is defined by two (or more) equations over a specified domain is called a piecewise function. Many Internet plans can be represented with piecewise functions. The graph of the piecewise function described above is shown in Figure 1.37.

Figure 1.37

The image shows a line graph for C of g.
Figure 1.37 Full Alternative Text

Example 6 Evaluating a Piecewise Function

Use the function that describes the Internet plan

C(g)={40if0g1540+0.60(g15)ifg>15

to find and interpret each of the following:

  1. C(10)

  2. C(45).

Solution

  1. To find C(10), we let g=10. Because 10 lies between 0 and 15, we use the first line of the piecewise function.

    C(g)=40This is the function's equation for 0g15.C(10)=40Replace g with 10. Regardless of this functionsinput, the constant output is 40.

    This means that with 10 GB of data, the monthly cost is $40. This can be visually represented by the point (10, 40) on the first piece of the graph in Figure 1.37.

  2. To find C(45), we let g=45. Because 45 is greater than 15, we use the second line of the piecewise function.

    C(g)=40+0.60(g15)This is the functionequation for g>15.  C(45)=40+0.60(4515)Replace  g  with 45.=40+0.60(30)Subtract within parentheses: 4515=30.=40+18Multiply: 0.60(30)=18.=58Add: 40+18=58.

    Thus, C(45)=58. This means that with 45 GB of data, the monthly cost is $58. This can be visually represented by the point (45, 58) on the second piece of the graph in Figure 1.37.

Check Point 6

  • Use the function in Example 6 to find and interpret each of the following:

    1. C(5)

    2. C(50).

    Identify your solutions by points on the graph in Figure 1.37.

Example 7 Graphing a Piecewise Function

Graph the piecewise function defined by

f(x)={x+2ifx1.4+xifx>1.

Solution

We graph f in two parts, using a partial table of coordinates to create each piece. The tables of coordinates and the completed graph are shown in Figure 1.38.

Figure 1.38Graphing a piecewise function

A graph plots a horizontal line that extends from an open point at (1, 4), and another line that falls from a closed point at (1, 3) through (negative 2, 0).

We can use the graph of the piecewise function in Figure 1.38 to find the range of f. The range of the blue piece on the left is {y|y3}. The range of the red horizontal piece on the right is {y|y=4}. Thus, the range of f is

{y|y3}{y|y=4}, or (, 3]{4}.

Check Point 7

  • Graph the piecewise function defined by

    f(x)={3ifx1x2ifx>1.

Some piecewise functions are called step functions because their graphs form discontinuous steps. One such function is called the greatest integer function, symbolized by int(x) or x, where

int(x)=the greatest integer that is less than or equal tox.

For example,

int of 1 = 1 comma int of 1.3 = 1 comma int of 1.5 = 1 comma int of 1.9 = 1. A note reads, 1 is the greatest integer that is less than or equal to 1, 1.3, 1.5, and 1.9.

Here are some additional examples:

int of 2 = 2 comma int of 2.3 = 2 comma int of 2.5 = 2 comma int of 2.9 = 2. A note reads, 2 is the greatest integer that is less than or equal to 1, 2.3, 2.5, and 2.9.

Notice how we jumped from 1 to 2 in the function values for int(x). In particular,

If1x<2, thenint(x)=1.If2x<3, thenint(x)=2.

The graph of f(x)=int(x) is shown in Figure 1.39. The graph of the greatest integer function jumps vertically one unit at each integer. However, the graph is constant between each pair of consecutive integers. The rightmost horizontal step shown in the graph illustrates that

If5x<6, thenint(x)=5.

Figure 1.39The graph of the greatest integer function

The graph of f of x = int of x plots a step function with a set of closed points and their corresponding open points.
Figure 1.39 Full Alternative Text

In general,

Ifnx<n+1, wheren is an integer, thenint(x)=n.
Objective 5: Understand and use piecewise functions

Piecewise Functions

  1. Objective 5 Understand and use piecewise functions.

Watch Video

A cellphone company offers the following plan for a mobile hotspot:

We can represent this plan mathematically by writing the total monthly cost, C, as a function of the amount of data used, g.

C of g = 40, if g is greater than or equal to 0 and is less than or equal to 15 and 40 + 0.60 left parenthesis g minus 15 right parenthesis, if g is greater than 15.
1.3-148 Full Alternative Text

A function that is defined by two (or more) equations over a specified domain is called a piecewise function. Many Internet plans can be represented with piecewise functions. The graph of the piecewise function described above is shown in Figure 1.37.

Figure 1.37

The image shows a line graph for C of g.
Figure 1.37 Full Alternative Text

Example 6 Evaluating a Piecewise Function

Use the function that describes the Internet plan

C(g)={40if0g1540+0.60(g15)ifg>15

to find and interpret each of the following:

  1. C(10)

  2. C(45).

Solution

  1. To find C(10), we let g=10. Because 10 lies between 0 and 15, we use the first line of the piecewise function.

    C(g)=40This is the function's equation for 0g15.C(10)=40Replace g with 10. Regardless of this functionsinput, the constant output is 40.

    This means that with 10 GB of data, the monthly cost is $40. This can be visually represented by the point (10, 40) on the first piece of the graph in Figure 1.37.

  2. To find C(45), we let g=45. Because 45 is greater than 15, we use the second line of the piecewise function.

    C(g)=40+0.60(g15)This is the functionequation for g>15.  C(45)=40+0.60(4515)Replace  g  with 45.=40+0.60(30)Subtract within parentheses: 4515=30.=40+18Multiply: 0.60(30)=18.=58Add: 40+18=58.

    Thus, C(45)=58. This means that with 45 GB of data, the monthly cost is $58. This can be visually represented by the point (45, 58) on the second piece of the graph in Figure 1.37.

Check Point 6

  • Use the function in Example 6 to find and interpret each of the following:

    1. C(5)

    2. C(50).

    Identify your solutions by points on the graph in Figure 1.37.

Example 7 Graphing a Piecewise Function

Graph the piecewise function defined by

f(x)={x+2ifx1.4+xifx>1.

Solution

We graph f in two parts, using a partial table of coordinates to create each piece. The tables of coordinates and the completed graph are shown in Figure 1.38.

Figure 1.38Graphing a piecewise function

A graph plots a horizontal line that extends from an open point at (1, 4), and another line that falls from a closed point at (1, 3) through (negative 2, 0).

We can use the graph of the piecewise function in Figure 1.38 to find the range of f. The range of the blue piece on the left is {y|y3}. The range of the red horizontal piece on the right is {y|y=4}. Thus, the range of f is

{y|y3}{y|y=4}, or (, 3]{4}.

Check Point 7

  • Graph the piecewise function defined by

    f(x)={3ifx1x2ifx>1.

Some piecewise functions are called step functions because their graphs form discontinuous steps. One such function is called the greatest integer function, symbolized by int(x) or x, where

int(x)=the greatest integer that is less than or equal tox.

For example,

int of 1 = 1 comma int of 1.3 = 1 comma int of 1.5 = 1 comma int of 1.9 = 1. A note reads, 1 is the greatest integer that is less than or equal to 1, 1.3, 1.5, and 1.9.

Here are some additional examples:

int of 2 = 2 comma int of 2.3 = 2 comma int of 2.5 = 2 comma int of 2.9 = 2. A note reads, 2 is the greatest integer that is less than or equal to 1, 2.3, 2.5, and 2.9.

Notice how we jumped from 1 to 2 in the function values for int(x). In particular,

If1x<2, thenint(x)=1.If2x<3, thenint(x)=2.

The graph of f(x)=int(x) is shown in Figure 1.39. The graph of the greatest integer function jumps vertically one unit at each integer. However, the graph is constant between each pair of consecutive integers. The rightmost horizontal step shown in the graph illustrates that

If5x<6, thenint(x)=5.

Figure 1.39The graph of the greatest integer function

The graph of f of x = int of x plots a step function with a set of closed points and their corresponding open points.
Figure 1.39 Full Alternative Text

In general,

Ifnx<n+1, wheren is an integer, thenint(x)=n.
Objective 6: Find and simplify a function’s difference quotient

Functions and Difference Quotients

  1. Objective 6 Find and simplify a function’s difference quotient.

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In Section 1.5, we will be studying the average rate of change of a function. A ratio, called the difference quotient, plays an important role in understanding the rate at which functions change.

Definition of the Difference Quotient of a Function

The expression

f(x+h)f(x)h

for h0 is called the difference quotient of the function f.

Example 8 Evaluating and Simplifying a Difference Quotient

If f(x)=2x2x+3, find and simplify each expression:

  1. f(x+h)

  2. f(x+h)f(x)h, h0.

Solution

  1. We find f(x+h) by replacing x with x+h each time that x appears in the equation.

    The image shows a step of a mathematical expression to expand f of x = 2 x squared minus x + 3.

  2. Using our result from part (a), we obtain the following:

    Simplification of an expression, start fraction f of start expression x + h end expression minus f of x over h end fraction.

Check Point 8

  • If f(x)=2x2+x+5, find and simplify each expression:

    1. f(x+h)

    2. f(x+h)f(x)h, h0.

1.3: Exercise Set

1.3 Exercise Set

Practice Exercises

In Exercises 112, use the graph to determine

  1. intervals on which the function is increasing, if any.

  2. intervals on which the function is decreasing, if any.

  3. intervals on which the function is constant, if any.

  1. 1.

    A graph is an upward opening parabola with indefinite ends that falls from (negative 4, 5) through (negative 3, 0) to its vertex at (negative 1, negative 4), and then rises through (1, 0) to point (4, 4).
  2. 2.

    A graph is a downward opening parabola with indefinite ends that rises from (negative 4, negative 5) through (negative 3, 0) to its vertex at (negative 1, 4) and falls through (1, 0) to point (2, negative 5). All values are estimated.
  3. 3.

    A graph is a curve with a closed point and indefinite end at (0, 1) and passes through (4, 3). All values are estimated.
  4. 4.

    A graph is a curve with a closed point and indefinite end that starts at (negative 1, 0) and passes through (0, 1) and (3, 2). All values are estimated.
  5. 5.

    A graph is a line with closed points that falls from (negative 2, 6), passes through (0, 4), (4, 0), (0, 4) and ends at (6, negative 2). All values are estimated.
  6. 6.

    A graph is a line with closed points that rises from (negative 3, negative 5) (negative 0.5, 0), passes through (0, 1), and ends at (2, 5). All values are estimated.
  7. 7.

    A graph is a line that rises through (negative 4, negative 5) to (negative 1, 2) and then extends horizontally through (0, negative 2) and (4, negative 2). All values are estimated.
  8. 8.

    A graph is a line that passes along the negative x axis to the origin and then rises through (1, 2) and (2, 4). All values are estimated.
  9. 9.

    A graph is an m shaped curve with indefinite ends that start in the third quadrant and end in the fourth quadrant.
  10. 10.

    A graph is a sinusoidal curve with closed ends that starts at (negative 5, 0), passes through (negative 4, 4), (negative 3, 0), (negative 2, negative 4), (negative 1, 0), (0, 4), (1, 0), (2, negative 4), (3, 0), (4, 4), and ends at (5, 0).
  11. 11.

    A graph is a line curve with indefinite ends that passes through (negative 3, negative 2), (negative 2, negative 2), (0, negative 1), (2, 0), (4, 2), and (5, 2).
  12. 12.

    A graph is a curve with indefinite ends that passes through (negative 5, 2), (negative 4, 2), (0, 0), (2, negative 2), and (3, negative 2).

In Exercises 1316, the graph of a function f is given. Use the graph to find each of the following:

  1. The numbers, if any, at which f has a relative maximum. What are these relative maxima?

  2. The numbers, if any, at which f has a relative minimum. What are these relative minima?

  1. 13.

    A graph is a w shaped curve with indefinite ends that passes through (negative 4, 3), (negative 3, 0), (0, 4), (3, 0), and (4, 3).
  2. 14.

    A graph is a curve that passes through (negative 3, negative 1), (0, 2), and (3, negative 1).
  3. 15.

    A graphing calculator screen displays a N shaped curve, f of x = 2 x cubed + 3 x squared minus 12 x + 1, which passes through the third quadrant, second quadrant, fourth quadrant, and then first quadrant.

  4. 16.

    A graphing calculator screen displays a curve, f of x = 2 x cubed minus 15 x squared + 24 x + 19, which passes through the third, second, and first quadrant.

In Exercises 1732, determine whether the graph of each equation is symmetric with respect to the y-axis, the x-axis, the origin, more than one of these, or none of these.

  1. 17. y=x2+6

  2. 18. y=x22

  3. 19. x=y2+6

  4. 20. x=y22

  5. 21. y2=x2+6

  6. 22. y2=x22

  7. 23. y=2x+3

  8. 24. y=2x+5

  9. 25. x2y3=2

  10. 26. x3y2=5

  11. 27. x2+y2=100

  12. 28. x2+y2=49

  13. 29. x2y2+3xy=1

  14. 30. x2y2+5xy=2

  15. 31. y4=x3+6

  16. 32. y5=x4+2

In Exercises 3336, use possible symmetry to determine whether each graph is the graph of an even function, an odd function, or a function that is neither even nor odd.

  1. 33.

    A graph is a curve with indefinite ends that passes through (negative 2, 4 fifths), (0, 4), and (2, 4 fifths).
  2. 34.

    A graph plots two curves with indefinite ends. The first curve passes through (1, 4), (2, 2), and (4, 1). The second curve passes through (negative 4, negative 1), (negative 2, negative 2), and (negative 1, negative 4).
  3. 35.

    A graph is a curve that passes through (negative 1, 1), (0, 0), and (1, negative 1).
  4. 36.

    A graph is a curve that passes through (negative 1, 3), (0, 2), (1, 1), and (2, negative 5).

In Exercises 3748, determine whether each function is even, odd, or neither. Then determine whether the function’s graph is symmetric with respect to the y-axis, the origin, or neither.

  1. 37. f(x)=x3+x

  2. 38. f(x)=x3x

  3. 39. g(x)=x2+x

  4. 40. g(x)=x2x

  5. 41. h(x)=x2x4

  6. 42. h(x)=2x2+x4

  7. 43. f(x)=x2x4+1

  8. 44. f(x)=2x2+x4+1

  9. 45. f(x)=15 x63x2

  10. 46. f(x)=2x36x5

  11. 47. f(x)=x1x2

  12. 48. f(x)=x21x2

  13. 49. Use the graph of f to determine each of the following. Where applicable, use interval notation.

    The graph of y = f of x is a horizontal line with an indefinite end from (negative 1, 4) to (0, 4) and a curve with an indefinite end that passes through (0, 4), (1, 0), (4, negative 4), and (7, 0).
    1. the domain of f

    2. the range of f

    3. the x-intercepts

    4. the y-intercept

    5. intervals on which f is increasing

    6. intervals on which f is decreasing

    7. intervals on which f is constant

    8. the number at which f has a relative minimum

    9. the relative minimum of f

    10. f(3)

    11. the values of x for which f(x)=2

    12. Is f even, odd, or neither?

  14. 50. Use the graph of f to determine each of the following. Where applicable, use interval notation.

    The graph of y = f of x is a curve with indefinite ends that passes through (negative 4, 0), (negative 2, 4), (0, 1), (3, 2), and (4, 0).
    1. the domain of f

    2. the range of f

    3. the x-intercepts

    4. the y-intercept

    5. intervals on which f is increasing

    6. intervals on which f is decreasing

    7. values of x for which f(x)0

    8. the numbers at which f has a relative maximum

    9. the relative maxima of f

    10. f(2)

    11. the values of x for which f(x)=0

    12. Is f even, odd, or neither?

  15. 51. Use the graph of f to determine each of the following. Where applicable, use interval notation.

    The graph of y = f of x is a curve with indefinite ends that starts at (3, 0), passes through (1, 4), (0, 3), and (negative 3, 0).
    1. the domain of f

    2. the range of f

    3. the zeros of f

    4. f(0)

    5. intervals on which f is increasing

    6. intervals on which f is decreasing

    7. values of x for which f(x)0

    8. any relative maxima and the numbers at which they occur

    9. the value of x for which f(x)=4

    10. Is f(1) positive or negative?

  16. 52. Use the graph of f to determine each of the following. Where applicable, use interval notation.

    The graph of y = f of x is a line that rises through (negative 5, negative 2) to (negative 2, 1), a horizontal line from (negative 2, 1) to (2, 1), and a line that falls to a closed point (5, negative 2).
    1. the domain of f

    2. the range of f

    3. the zeros of f

    4. f(0)

    5. intervals on which f is increasing

    6. intervals on which f is decreasing

    7. intervals on which f is constant

    8. values of x for which f(x)>0

    9. values of x for which f(x)=2

    10. Is f(4) positive or negative?

    11. Is f even, odd, or neither?

    12. Is f(2) a relative maximum?

In Exercises 5358, evaluate each piecewise function at the given values of the independent variable.

  1. 53. f(x)={3x+5ifx<04x+7ifx0

    1. f(2)

    2. f(0)

    3. f(3)

  2. 54. f(x)={6x1ifx<07x+3ifx0

    1. f(3)

    2. f(0)

    3. f(4)

  3. 55. g(x)={x+3ifx3(x+3)ifx<3

    1. g(0)

    2. g(6)

    3. g(3)

  4. 56. g(x)={x+5ifx5(x+5)ifx<5

    1. g(0)

    2. g(6)

    3. g(5)

  5. 57. h(x)={x29x3ifx36ifx=3

    1. h(5)

    2. h(0)

    3. h(3)

  6. 58. h(x)={x225x5ifx510ifx=5

    1. h(7)

    2. h(0)

    3. h(5)

In Exercises 5970, the domain of each piecewise function is (, ).

  1. Graph each function.

  2. Use your graph to determine the function’s range.

  1. 59. f(x)={xifx<0xifx0

  2. 60. f(x)={xifx<0xifx0

  3. 61. f(x)={2xifx02ifx>0

  4. 62. f(x)={12 xifx03ifx>0

  5. 63. f(x)={x+3ifx<2x3ifx2

  6. 64. f(x)={x+2ifx<3x2ifx3

  7. 65. f(x)={3ifx13ifx>1

  8. 66. f(x)={4ifx14ifx>1

  9. 67. f(x)={12 x2ifx<12x1ifx1

  10. 68. f(x)={ 12 x2ifx<12x+1ifx1

  11. 69. f(x)={0ifx<4xif4x<0x2ifx0

  12. 70. f(x)={0ifx<3 xif3x<0x21ifx0

In Exercises 7192, find and simplify the difference quotient

f(x+h)f(x)h, h0

for the given function.

  1. 71. f(x)=4x

  2. 72. f(x)=7x

  3. 73. f(x)=3x+7

  4. 74. f(x)=6x+1

  5. 75. f(x)=x2

  6. 76. f(x)=2x2

  7. 77. f(x)=x24x+3

  8. 78. f(x)=x25x+8

  9. 79. f(x)=2x2+x1

  10. 80. f(x)=3x2+x+5

  11. 81. f(x)=x2+2x+4

  12. 82. f(x)=x23x+1

  13. 83. f(x)=2x2+5x+7

  14. 84. f(x)=3x2+2x1

  15. 85. f(x)=2x2x+3

  16. 86. f(x)=3x2+x1

  17. 87. f(x)=6

  18. 88. f(x)=7

  19. 89. f(x)=1x

  20. 90. f(x)=12x

  21. 91. f(x)=x

  22. 92. f(x)=x1

Practice PLUS

In Exercises 9394, let f be defined by the following graph:

The image shows a stepwise function graph.
  1. 93. Find

    f(1.5)+f(0.9)[f(π)]2+f(3)÷f(1)f(π).
  2. 94. Find

    f(2.5)f(1.9)[f(π)]2+f(3)÷f(1)f(π).

A cable company offers the following high-speed Internet plans. Also given are the piecewise functions that model these plans. Use this information to solve Exercises 9596.

Plan A

  • $40 per month includes 400 GB of data.

  • Additional data costs $0.20 per GB.

    C(g)={40if0g40040+0.20(g400)ifg>400

Plan B

  • $60 per month includes 1000 GB of data.

  • Additional data costs $0.20 per GB.

    C(g)={60if0g100060+0.20(g1000)ifg>1000
  1. 95. Simplify the algebraic expression in the second line of the piecewise function for plan A. Then use point-plotting to graph the function.

  2. 96. Simplify the algebraic expression in the second line of the piecewise function for plan B. Then use point-plotting to graph the function.

In Exercises 9798, write a piecewise function that models each high-speed Internet plan. Then graph the function.

  1. 97. $50 per month includes 600 GB. Additional data costs $0.30 per GB.

  2. 98. $80 per month includes 2000 GB. Additional data costs $0.35 per GB.

Application Exercises

With aging, body fat increases and muscle mass declines. The line graphs show the percent body fat in adult women and men as they age from 25 to 75 years. Use the graphs to solve Exercises 99106.

A graph plots the percent of body fat in adults.

Source: Thompson et al., The Science of Nutrition, Benjamin Cummings, 2008

  1. 99. State the intervals on which the graph giving the percent body fat in women is increasing and decreasing.

  2. 100. State the intervals on which the graph giving the percent body fat in men is increasing and decreasing.

  3. 101. For what age does the percent body fat in women reach a maximum? What is the percent body fat for that age?

  4. 102. At what age does the percent body fat in men reach a maximum? What is the percent body fat for that age?

  5. 103. Use interval notation to give the domain and the range for the graph of the function for women.

  6. 104. Use interval notation to give the domain and the range for the graph of the function for men.

  7. 105. The function p(x)=0.002x2+0.15x+22.86 models percent body fat, p(x), where x is the number of years a person’s age exceeds 25. Use the graphs to determine whether this model describes percent body fat in women or in men.

  8. 106. The function p(x)=0.004x2+0.25x+33.64 models percent body fat, p(x), where x is the number of years a person’s age exceeds 25. Use the graphs to determine whether this model describes percent body fat in women or in men.

Here is the Federal Tax Rate Schedule X that specifies the tax owed by a single taxpayer for 2020.

If Your Taxable Income Is Over But Not Over The Tax You Owe Is Of the Amount Over
$          0 $    9,875                       10% $          0
$    9875 $  40,125 $   987.50+12% $    9875
$  40,125 $  85,525 $   4617.50+22% $  40,125
$  85,525 $163,300 $14,605.50+24% $  85,525
$163,300 $207,350 $33,271.50+32% $163,300
$207,350 $518,400 $47,367.50+35% $207,350
$518,400       − $   156,235+37% $518,400

The preceding tax table can be modeled by a piecewise function, where x represents the taxable income of a single taxpayer and T(x) is the tax owed:

T(x)={0.10xif0<x9875987.50+0.12(x9875)if9875<x40,1254617.50+0.22(x40,125)if40,125<x85,52514,605.50+0.24(x85,525)if85,525<x163,30033,271.50+0.32(x163,300)if163,300<x207,350wewdwdwdss?sdwdwdwddw¯if207,350<x518,400sssasassxxax?xssceddvdddd¯ifx>518,400

Use this information to solve Exercises 107108.

  1. 107. Find and interpret T(35,000).

  2. 108. Find and interpret T(50,000).

In Exercises 109110, refer to the preceding tax table.

  1. 109. Find the algebraic expression for the missing piece of T(x) that models tax owed for the domain (207,350,518,400].

  2. 110. Find the algebraic expression for the missing piece of T(x) that models tax owed for the domain (518,400, ).

The figure shows the cost of mailing a first-class letter, f(x), as a function of its weight, x, in ounces, in June 2020. Use the graph to solve Exercises 111114.

The image shows a function graph that plots cost, in dollars, versus weight, in ounces.

Source: Lynn E. Baring, Postmaster, Inverness, CA

  1. 111. Find f(3). What does this mean in terms of the variables in this situation?

  2. 112. Find f(3.75). What does this mean in terms of the variables in this situation?

  3. 113. What is the cost of mailing a letter that weighs 1.5 ounces?

  4. 114. What is the cost of mailing a letter that weighs 1.8 ounces?

  5. 115. Furry Finances. A pet insurance policy has a monthly rate that is a function of the age of the insured dog or cat. For pets whose age does not exceed 4, the monthly cost is $20. The cost then increases by $2 for each successive year of the pet’s age.

    Age Not Exceeding Monthly Cost
    4 $20
    5 $22
    6 $24

    The cost schedule continues in this manner for ages not exceeding 10. The cost for pets whose ages exceed 10 is $40. Use this information to create a graph that shows the monthly cost of the insurance, f(x), for a pet of age x, where the function’s domain is [0, 14].

Explaining the Concepts

  1. 116. What does it mean if a function f is increasing on an interval?

  2. 117. Suppose that a function f whose graph contains no breaks or gaps on (a, c) is increasing on (a, b), decreasing on (b, c), and defined at b. Describe what occurs at x=b. What does the function value f(b) represent?

  3. 118. Given an equation in x and y, how do you determine if its graph is symmetric with respect to the y-axis?

  4. 119. Given an equation in x and y, how do you determine if its graph is symmetric with respect to the x-axis?

  5. 120. Given an equation in x and y, how do you determine if its graph is symmetric with respect to the origin?

  6. 121. If you are given a function’s graph, how do you determine if the function is even, odd, or neither?

  7. 122. If you are given a function’s equation, how do you determine if the function is even, odd, or neither?

  8. 123. What is a piecewise function?

  9. 124. Explain how to find the difference quotient of a function f, f(x+h)f(x)h, if an equation for f is given.

Technology Exercises

  1. 125. The function

    f(x)=0.00002x3+0.008x20.3x+6.95

    models the number of annual physician visits, f(x), by a person of age x. Graph the function in a [0,100,5] by [0,40,2] viewing rectangle. What does the shape of the graph indicate about the relationship between one’s age and the number of annual physician visits? Use the  TABLE  or minimum function capability to find the coordinates of the minimum point on the graph of the function. What does this mean?

In Exercises 126131, use a graphing utility to graph each function. Use a [5, 5, 1] by [5, 5, 1] viewing rectangle. Then find the intervals on which the function is increasing, decreasing, or constant.

  1. 126. f(x)=x36x2+9x+1

  2. 127. g(x)=|4x2|

  3. 128. h(x)=|x2|+|x+2|

  4. 129. f(x)=x13(x4)

  5. 130. g(x)=x23

  6. 131. h(x)=2x25

  7. 132.

    1. Graph the functions f(x)=xn for n=2, 4, and 6 in a [2, 2, 1] by [1, 3, 1] viewing rectangle.

    2. Graph the functions f(x)=xn for n=1, 3, and 5 in a [2, 2, 1] by [2, 2, 1] viewing rectangle.

    3. If n is positive and even, where is the graph of f(x)=xn increasing and where is it decreasing?

    4. If n is positive and odd, what can you conclude about the graph of f(x)=xn in terms of increasing or decreasing behavior?

    5. Graph all six functions in a [1, 3, 1] by [1, 3, 1] viewing rectangle. What do you observe about the graphs in terms of how flat or how steep they are?

Critical Thinking Exercises

Make Sense? In Exercises 133136, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 133. My graph is decreasing on (, a) and increasing on (a, ), so f(a) must be a relative maximum.

  2. 134. This work by artist Scott Kim has the same kind of symmetry as an even function.

    The word Dyslexia is written such that the alphabets on either side of the letter, l, appear as a mirror image of each other. Double prime DYSLEXIA comma double prime 19 81.
  3. 135. I graphed

    f(x)={2ifx43ifx=4

    and one piece of my graph is a single point.

  4. 136. I noticed that the difference quotient is always zero if f(x)=c, where c is any constant.

  5. 137. Sketch the graph of f using the following properties. (More than one correct graph is possible.) f is a piecewise function that is decreasing on (, 2), f(2)=0, f is increasing on (2, ), and the range of f is [0, ).

  6. 138. Define a piecewise function on the intervals (, 2], (2, 5), and [5, ) that does not “jump” at 2 or 5 such that one piece is a constant function, another piece is an increasing function, and the third piece is a decreasing function.

  7. 139. Suppose that h(x)=f(x)g(x). The function f can be even, odd, or neither. The same is true for the function g.

    1. Under what conditions is h definitely an even function?

    2. Under what conditions is h definitely an odd function?

Retaining the Concepts

  1. 140. You invested $80,000 in two accounts paying 1.5% and 1.7% annual interest. If the total interest earned for the year was $1320, how much was invested at each rate? (Section P.8, Example 5)

  2. 141. Solve for A: C=A+Ar.

    (Section P.7, Example 5)

  3. 142. Solve by the quadratic formula: 5x26x8=0.

    (Section P.7, Example 10)

Preview Exercises

Exercises 143145 will help you prepare for the material covered in the next section.

  1. 143. If (x1, y1)=(3, 1) and (x2, y2)=(2, 4), find y2y1x2x1.

  2. 144. Find the ordered pairs (_______, 0) and (0, _______) satisfying 4x3y6=0.

  3. 145. Solve for y: 3x+2y4=0.

1.3: Exercise Set

1.3 Exercise Set

Practice Exercises

In Exercises 112, use the graph to determine

  1. intervals on which the function is increasing, if any.

  2. intervals on which the function is decreasing, if any.

  3. intervals on which the function is constant, if any.

  1. 1.

    A graph is an upward opening parabola with indefinite ends that falls from (negative 4, 5) through (negative 3, 0) to its vertex at (negative 1, negative 4), and then rises through (1, 0) to point (4, 4).
  2. 2.

    A graph is a downward opening parabola with indefinite ends that rises from (negative 4, negative 5) through (negative 3, 0) to its vertex at (negative 1, 4) and falls through (1, 0) to point (2, negative 5). All values are estimated.
  3. 3.

    A graph is a curve with a closed point and indefinite end at (0, 1) and passes through (4, 3). All values are estimated.
  4. 4.

    A graph is a curve with a closed point and indefinite end that starts at (negative 1, 0) and passes through (0, 1) and (3, 2). All values are estimated.
  5. 5.

    A graph is a line with closed points that falls from (negative 2, 6), passes through (0, 4), (4, 0), (0, 4) and ends at (6, negative 2). All values are estimated.
  6. 6.

    A graph is a line with closed points that rises from (negative 3, negative 5) (negative 0.5, 0), passes through (0, 1), and ends at (2, 5). All values are estimated.
  7. 7.

    A graph is a line that rises through (negative 4, negative 5) to (negative 1, 2) and then extends horizontally through (0, negative 2) and (4, negative 2). All values are estimated.
  8. 8.

    A graph is a line that passes along the negative x axis to the origin and then rises through (1, 2) and (2, 4). All values are estimated.
  9. 9.

    A graph is an m shaped curve with indefinite ends that start in the third quadrant and end in the fourth quadrant.
  10. 10.

    A graph is a sinusoidal curve with closed ends that starts at (negative 5, 0), passes through (negative 4, 4), (negative 3, 0), (negative 2, negative 4), (negative 1, 0), (0, 4), (1, 0), (2, negative 4), (3, 0), (4, 4), and ends at (5, 0).
  11. 11.

    A graph is a line curve with indefinite ends that passes through (negative 3, negative 2), (negative 2, negative 2), (0, negative 1), (2, 0), (4, 2), and (5, 2).
  12. 12.

    A graph is a curve with indefinite ends that passes through (negative 5, 2), (negative 4, 2), (0, 0), (2, negative 2), and (3, negative 2).

In Exercises 1316, the graph of a function f is given. Use the graph to find each of the following:

  1. The numbers, if any, at which f has a relative maximum. What are these relative maxima?

  2. The numbers, if any, at which f has a relative minimum. What are these relative minima?

  1. 13.

    A graph is a w shaped curve with indefinite ends that passes through (negative 4, 3), (negative 3, 0), (0, 4), (3, 0), and (4, 3).
  2. 14.

    A graph is a curve that passes through (negative 3, negative 1), (0, 2), and (3, negative 1).
  3. 15.

    A graphing calculator screen displays a N shaped curve, f of x = 2 x cubed + 3 x squared minus 12 x + 1, which passes through the third quadrant, second quadrant, fourth quadrant, and then first quadrant.

  4. 16.

    A graphing calculator screen displays a curve, f of x = 2 x cubed minus 15 x squared + 24 x + 19, which passes through the third, second, and first quadrant.

In Exercises 1732, determine whether the graph of each equation is symmetric with respect to the y-axis, the x-axis, the origin, more than one of these, or none of these.

  1. 17. y=x2+6

  2. 18. y=x22

  3. 19. x=y2+6

  4. 20. x=y22

  5. 21. y2=x2+6

  6. 22. y2=x22

  7. 23. y=2x+3

  8. 24. y=2x+5

  9. 25. x2y3=2

  10. 26. x3y2=5

  11. 27. x2+y2=100

  12. 28. x2+y2=49

  13. 29. x2y2+3xy=1

  14. 30. x2y2+5xy=2

  15. 31. y4=x3+6

  16. 32. y5=x4+2

In Exercises 3336, use possible symmetry to determine whether each graph is the graph of an even function, an odd function, or a function that is neither even nor odd.

  1. 33.

    A graph is a curve with indefinite ends that passes through (negative 2, 4 fifths), (0, 4), and (2, 4 fifths).
  2. 34.

    A graph plots two curves with indefinite ends. The first curve passes through (1, 4), (2, 2), and (4, 1). The second curve passes through (negative 4, negative 1), (negative 2, negative 2), and (negative 1, negative 4).
  3. 35.

    A graph is a curve that passes through (negative 1, 1), (0, 0), and (1, negative 1).
  4. 36.

    A graph is a curve that passes through (negative 1, 3), (0, 2), (1, 1), and (2, negative 5).

In Exercises 3748, determine whether each function is even, odd, or neither. Then determine whether the function’s graph is symmetric with respect to the y-axis, the origin, or neither.

  1. 37. f(x)=x3+x

  2. 38. f(x)=x3x

  3. 39. g(x)=x2+x

  4. 40. g(x)=x2x

  5. 41. h(x)=x2x4

  6. 42. h(x)=2x2+x4

  7. 43. f(x)=x2x4+1

  8. 44. f(x)=2x2+x4+1

  9. 45. f(x)=15 x63x2

  10. 46. f(x)=2x36x5

  11. 47. f(x)=x1x2

  12. 48. f(x)=x21x2

  13. 49. Use the graph of f to determine each of the following. Where applicable, use interval notation.

    The graph of y = f of x is a horizontal line with an indefinite end from (negative 1, 4) to (0, 4) and a curve with an indefinite end that passes through (0, 4), (1, 0), (4, negative 4), and (7, 0).
    1. the domain of f

    2. the range of f

    3. the x-intercepts

    4. the y-intercept

    5. intervals on which f is increasing

    6. intervals on which f is decreasing

    7. intervals on which f is constant

    8. the number at which f has a relative minimum

    9. the relative minimum of f

    10. f(3)

    11. the values of x for which f(x)=2

    12. Is f even, odd, or neither?

  14. 50. Use the graph of f to determine each of the following. Where applicable, use interval notation.

    The graph of y = f of x is a curve with indefinite ends that passes through (negative 4, 0), (negative 2, 4), (0, 1), (3, 2), and (4, 0).
    1. the domain of f

    2. the range of f

    3. the x-intercepts

    4. the y-intercept

    5. intervals on which f is increasing

    6. intervals on which f is decreasing

    7. values of x for which f(x)0

    8. the numbers at which f has a relative maximum

    9. the relative maxima of f

    10. f(2)

    11. the values of x for which f(x)=0

    12. Is f even, odd, or neither?

  15. 51. Use the graph of f to determine each of the following. Where applicable, use interval notation.

    The graph of y = f of x is a curve with indefinite ends that starts at (3, 0), passes through (1, 4), (0, 3), and (negative 3, 0).
    1. the domain of f

    2. the range of f

    3. the zeros of f

    4. f(0)

    5. intervals on which f is increasing

    6. intervals on which f is decreasing

    7. values of x for which f(x)0

    8. any relative maxima and the numbers at which they occur

    9. the value of x for which f(x)=4

    10. Is f(1) positive or negative?

  16. 52. Use the graph of f to determine each of the following. Where applicable, use interval notation.

    The graph of y = f of x is a line that rises through (negative 5, negative 2) to (negative 2, 1), a horizontal line from (negative 2, 1) to (2, 1), and a line that falls to a closed point (5, negative 2).
    1. the domain of f

    2. the range of f

    3. the zeros of f

    4. f(0)

    5. intervals on which f is increasing

    6. intervals on which f is decreasing

    7. intervals on which f is constant

    8. values of x for which f(x)>0

    9. values of x for which f(x)=2

    10. Is f(4) positive or negative?

    11. Is f even, odd, or neither?

    12. Is f(2) a relative maximum?

In Exercises 5358, evaluate each piecewise function at the given values of the independent variable.

  1. 53. f(x)={3x+5ifx<04x+7ifx0

    1. f(2)

    2. f(0)

    3. f(3)

  2. 54. f(x)={6x1ifx<07x+3ifx0

    1. f(3)

    2. f(0)

    3. f(4)

  3. 55. g(x)={x+3ifx3(x+3)ifx<3

    1. g(0)

    2. g(6)

    3. g(3)

  4. 56. g(x)={x+5ifx5(x+5)ifx<5

    1. g(0)

    2. g(6)

    3. g(5)

  5. 57. h(x)={x29x3ifx36ifx=3

    1. h(5)

    2. h(0)

    3. h(3)

  6. 58. h(x)={x225x5ifx510ifx=5

    1. h(7)

    2. h(0)

    3. h(5)

In Exercises 5970, the domain of each piecewise function is (, ).

  1. Graph each function.

  2. Use your graph to determine the function’s range.

  1. 59. f(x)={xifx<0xifx0

  2. 60. f(x)={xifx<0xifx0

  3. 61. f(x)={2xifx02ifx>0

  4. 62. f(x)={12 xifx03ifx>0

  5. 63. f(x)={x+3ifx<2x3ifx2

  6. 64. f(x)={x+2ifx<3x2ifx3

  7. 65. f(x)={3ifx13ifx>1

  8. 66. f(x)={4ifx14ifx>1

  9. 67. f(x)={12 x2ifx<12x1ifx1

  10. 68. f(x)={ 12 x2ifx<12x+1ifx1

  11. 69. f(x)={0ifx<4xif4x<0x2ifx0

  12. 70. f(x)={0ifx<3 xif3x<0x21ifx0

In Exercises 7192, find and simplify the difference quotient

f(x+h)f(x)h, h0

for the given function.

  1. 71. f(x)=4x

  2. 72. f(x)=7x

  3. 73. f(x)=3x+7

  4. 74. f(x)=6x+1

  5. 75. f(x)=x2

  6. 76. f(x)=2x2

  7. 77. f(x)=x24x+3

  8. 78. f(x)=x25x+8

  9. 79. f(x)=2x2+x1

  10. 80. f(x)=3x2+x+5

  11. 81. f(x)=x2+2x+4

  12. 82. f(x)=x23x+1

  13. 83. f(x)=2x2+5x+7

  14. 84. f(x)=3x2+2x1

  15. 85. f(x)=2x2x+3

  16. 86. f(x)=3x2+x1

  17. 87. f(x)=6

  18. 88. f(x)=7

  19. 89. f(x)=1x

  20. 90. f(x)=12x

  21. 91. f(x)=x

  22. 92. f(x)=x1

Practice PLUS

In Exercises 9394, let f be defined by the following graph:

The image shows a stepwise function graph.
  1. 93. Find

    f(1.5)+f(0.9)[f(π)]2+f(3)÷f(1)f(π).
  2. 94. Find

    f(2.5)f(1.9)[f(π)]2+f(3)÷f(1)f(π).

A cable company offers the following high-speed Internet plans. Also given are the piecewise functions that model these plans. Use this information to solve Exercises 9596.

Plan A

  • $40 per month includes 400 GB of data.

  • Additional data costs $0.20 per GB.

    C(g)={40if0g40040+0.20(g400)ifg>400

Plan B

  • $60 per month includes 1000 GB of data.

  • Additional data costs $0.20 per GB.

    C(g)={60if0g100060+0.20(g1000)ifg>1000
  1. 95. Simplify the algebraic expression in the second line of the piecewise function for plan A. Then use point-plotting to graph the function.

  2. 96. Simplify the algebraic expression in the second line of the piecewise function for plan B. Then use point-plotting to graph the function.

In Exercises 9798, write a piecewise function that models each high-speed Internet plan. Then graph the function.

  1. 97. $50 per month includes 600 GB. Additional data costs $0.30 per GB.

  2. 98. $80 per month includes 2000 GB. Additional data costs $0.35 per GB.

Application Exercises

With aging, body fat increases and muscle mass declines. The line graphs show the percent body fat in adult women and men as they age from 25 to 75 years. Use the graphs to solve Exercises 99106.

A graph plots the percent of body fat in adults.

Source: Thompson et al., The Science of Nutrition, Benjamin Cummings, 2008

  1. 99. State the intervals on which the graph giving the percent body fat in women is increasing and decreasing.

  2. 100. State the intervals on which the graph giving the percent body fat in men is increasing and decreasing.

  3. 101. For what age does the percent body fat in women reach a maximum? What is the percent body fat for that age?

  4. 102. At what age does the percent body fat in men reach a maximum? What is the percent body fat for that age?

  5. 103. Use interval notation to give the domain and the range for the graph of the function for women.

  6. 104. Use interval notation to give the domain and the range for the graph of the function for men.

  7. 105. The function p(x)=0.002x2+0.15x+22.86 models percent body fat, p(x), where x is the number of years a person’s age exceeds 25. Use the graphs to determine whether this model describes percent body fat in women or in men.

  8. 106. The function p(x)=0.004x2+0.25x+33.64 models percent body fat, p(x), where x is the number of years a person’s age exceeds 25. Use the graphs to determine whether this model describes percent body fat in women or in men.

Here is the Federal Tax Rate Schedule X that specifies the tax owed by a single taxpayer for 2020.

If Your Taxable Income Is Over But Not Over The Tax You Owe Is Of the Amount Over
$          0 $    9,875                       10% $          0
$    9875 $  40,125 $   987.50+12% $    9875
$  40,125 $  85,525 $   4617.50+22% $  40,125
$  85,525 $163,300 $14,605.50+24% $  85,525
$163,300 $207,350 $33,271.50+32% $163,300
$207,350 $518,400 $47,367.50+35% $207,350
$518,400       − $   156,235+37% $518,400

The preceding tax table can be modeled by a piecewise function, where x represents the taxable income of a single taxpayer and T(x) is the tax owed:

T(x)={0.10xif0<x9875987.50+0.12(x9875)if9875<x40,1254617.50+0.22(x40,125)if40,125<x85,52514,605.50+0.24(x85,525)if85,525<x163,30033,271.50+0.32(x163,300)if163,300<x207,350wewdwdwdss?sdwdwdwddw¯if207,350<x518,400sssasassxxax?xssceddvdddd¯ifx>518,400

Use this information to solve Exercises 107108.

  1. 107. Find and interpret T(35,000).

  2. 108. Find and interpret T(50,000).

In Exercises 109110, refer to the preceding tax table.

  1. 109. Find the algebraic expression for the missing piece of T(x) that models tax owed for the domain (207,350,518,400].

  2. 110. Find the algebraic expression for the missing piece of T(x) that models tax owed for the domain (518,400, ).

The figure shows the cost of mailing a first-class letter, f(x), as a function of its weight, x, in ounces, in June 2020. Use the graph to solve Exercises 111114.

The image shows a function graph that plots cost, in dollars, versus weight, in ounces.

Source: Lynn E. Baring, Postmaster, Inverness, CA

  1. 111. Find f(3). What does this mean in terms of the variables in this situation?

  2. 112. Find f(3.75). What does this mean in terms of the variables in this situation?

  3. 113. What is the cost of mailing a letter that weighs 1.5 ounces?

  4. 114. What is the cost of mailing a letter that weighs 1.8 ounces?

  5. 115. Furry Finances. A pet insurance policy has a monthly rate that is a function of the age of the insured dog or cat. For pets whose age does not exceed 4, the monthly cost is $20. The cost then increases by $2 for each successive year of the pet’s age.

    Age Not Exceeding Monthly Cost
    4 $20
    5 $22
    6 $24

    The cost schedule continues in this manner for ages not exceeding 10. The cost for pets whose ages exceed 10 is $40. Use this information to create a graph that shows the monthly cost of the insurance, f(x), for a pet of age x, where the function’s domain is [0, 14].

Explaining the Concepts

  1. 116. What does it mean if a function f is increasing on an interval?

  2. 117. Suppose that a function f whose graph contains no breaks or gaps on (a, c) is increasing on (a, b), decreasing on (b, c), and defined at b. Describe what occurs at x=b. What does the function value f(b) represent?

  3. 118. Given an equation in x and y, how do you determine if its graph is symmetric with respect to the y-axis?

  4. 119. Given an equation in x and y, how do you determine if its graph is symmetric with respect to the x-axis?

  5. 120. Given an equation in x and y, how do you determine if its graph is symmetric with respect to the origin?

  6. 121. If you are given a function’s graph, how do you determine if the function is even, odd, or neither?

  7. 122. If you are given a function’s equation, how do you determine if the function is even, odd, or neither?

  8. 123. What is a piecewise function?

  9. 124. Explain how to find the difference quotient of a function f, f(x+h)f(x)h, if an equation for f is given.

Technology Exercises

  1. 125. The function

    f(x)=0.00002x3+0.008x20.3x+6.95

    models the number of annual physician visits, f(x), by a person of age x. Graph the function in a [0,100,5] by [0,40,2] viewing rectangle. What does the shape of the graph indicate about the relationship between one’s age and the number of annual physician visits? Use the  TABLE  or minimum function capability to find the coordinates of the minimum point on the graph of the function. What does this mean?

In Exercises 126131, use a graphing utility to graph each function. Use a [5, 5, 1] by [5, 5, 1] viewing rectangle. Then find the intervals on which the function is increasing, decreasing, or constant.

  1. 126. f(x)=x36x2+9x+1

  2. 127. g(x)=|4x2|

  3. 128. h(x)=|x2|+|x+2|

  4. 129. f(x)=x13(x4)

  5. 130. g(x)=x23

  6. 131. h(x)=2x25

  7. 132.

    1. Graph the functions f(x)=xn for n=2, 4, and 6 in a [2, 2, 1] by [1, 3, 1] viewing rectangle.

    2. Graph the functions f(x)=xn for n=1, 3, and 5 in a [2, 2, 1] by [2, 2, 1] viewing rectangle.

    3. If n is positive and even, where is the graph of f(x)=xn increasing and where is it decreasing?

    4. If n is positive and odd, what can you conclude about the graph of f(x)=xn in terms of increasing or decreasing behavior?

    5. Graph all six functions in a [1, 3, 1] by [1, 3, 1] viewing rectangle. What do you observe about the graphs in terms of how flat or how steep they are?

Critical Thinking Exercises

Make Sense? In Exercises 133136, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 133. My graph is decreasing on (, a) and increasing on (a, ), so f(a) must be a relative maximum.

  2. 134. This work by artist Scott Kim has the same kind of symmetry as an even function.

    The word Dyslexia is written such that the alphabets on either side of the letter, l, appear as a mirror image of each other. Double prime DYSLEXIA comma double prime 19 81.
  3. 135. I graphed

    f(x)={2ifx43ifx=4

    and one piece of my graph is a single point.

  4. 136. I noticed that the difference quotient is always zero if f(x)=c, where c is any constant.

  5. 137. Sketch the graph of f using the following properties. (More than one correct graph is possible.) f is a piecewise function that is decreasing on (, 2), f(2)=0, f is increasing on (2, ), and the range of f is [0, ).

  6. 138. Define a piecewise function on the intervals (, 2], (2, 5), and [5, ) that does not “jump” at 2 or 5 such that one piece is a constant function, another piece is an increasing function, and the third piece is a decreasing function.

  7. 139. Suppose that h(x)=f(x)g(x). The function f can be even, odd, or neither. The same is true for the function g.

    1. Under what conditions is h definitely an even function?

    2. Under what conditions is h definitely an odd function?

Retaining the Concepts

  1. 140. You invested $80,000 in two accounts paying 1.5% and 1.7% annual interest. If the total interest earned for the year was $1320, how much was invested at each rate? (Section P.8, Example 5)

  2. 141. Solve for A: C=A+Ar.

    (Section P.7, Example 5)

  3. 142. Solve by the quadratic formula: 5x26x8=0.

    (Section P.7, Example 10)

Preview Exercises

Exercises 143145 will help you prepare for the material covered in the next section.

  1. 143. If (x1, y1)=(3, 1) and (x2, y2)=(2, 4), find y2y1x2x1.

  2. 144. Find the ordered pairs (_______, 0) and (0, _______) satisfying 4x3y6=0.

  3. 145. Solve for y: 3x+2y4=0.

1.3: Exercise Set

1.3 Exercise Set

Practice Exercises

In Exercises 112, use the graph to determine

  1. intervals on which the function is increasing, if any.

  2. intervals on which the function is decreasing, if any.

  3. intervals on which the function is constant, if any.

  1. 1.

    A graph is an upward opening parabola with indefinite ends that falls from (negative 4, 5) through (negative 3, 0) to its vertex at (negative 1, negative 4), and then rises through (1, 0) to point (4, 4).
  2. 2.

    A graph is a downward opening parabola with indefinite ends that rises from (negative 4, negative 5) through (negative 3, 0) to its vertex at (negative 1, 4) and falls through (1, 0) to point (2, negative 5). All values are estimated.
  3. 3.

    A graph is a curve with a closed point and indefinite end at (0, 1) and passes through (4, 3). All values are estimated.
  4. 4.

    A graph is a curve with a closed point and indefinite end that starts at (negative 1, 0) and passes through (0, 1) and (3, 2). All values are estimated.
  5. 5.

    A graph is a line with closed points that falls from (negative 2, 6), passes through (0, 4), (4, 0), (0, 4) and ends at (6, negative 2). All values are estimated.
  6. 6.

    A graph is a line with closed points that rises from (negative 3, negative 5) (negative 0.5, 0), passes through (0, 1), and ends at (2, 5). All values are estimated.
  7. 7.

    A graph is a line that rises through (negative 4, negative 5) to (negative 1, 2) and then extends horizontally through (0, negative 2) and (4, negative 2). All values are estimated.
  8. 8.

    A graph is a line that passes along the negative x axis to the origin and then rises through (1, 2) and (2, 4). All values are estimated.
  9. 9.

    A graph is an m shaped curve with indefinite ends that start in the third quadrant and end in the fourth quadrant.
  10. 10.

    A graph is a sinusoidal curve with closed ends that starts at (negative 5, 0), passes through (negative 4, 4), (negative 3, 0), (negative 2, negative 4), (negative 1, 0), (0, 4), (1, 0), (2, negative 4), (3, 0), (4, 4), and ends at (5, 0).
  11. 11.

    A graph is a line curve with indefinite ends that passes through (negative 3, negative 2), (negative 2, negative 2), (0, negative 1), (2, 0), (4, 2), and (5, 2).
  12. 12.

    A graph is a curve with indefinite ends that passes through (negative 5, 2), (negative 4, 2), (0, 0), (2, negative 2), and (3, negative 2).

In Exercises 1316, the graph of a function f is given. Use the graph to find each of the following:

  1. The numbers, if any, at which f has a relative maximum. What are these relative maxima?

  2. The numbers, if any, at which f has a relative minimum. What are these relative minima?

  1. 13.

    A graph is a w shaped curve with indefinite ends that passes through (negative 4, 3), (negative 3, 0), (0, 4), (3, 0), and (4, 3).
  2. 14.

    A graph is a curve that passes through (negative 3, negative 1), (0, 2), and (3, negative 1).
  3. 15.

    A graphing calculator screen displays a N shaped curve, f of x = 2 x cubed + 3 x squared minus 12 x + 1, which passes through the third quadrant, second quadrant, fourth quadrant, and then first quadrant.

  4. 16.

    A graphing calculator screen displays a curve, f of x = 2 x cubed minus 15 x squared + 24 x + 19, which passes through the third, second, and first quadrant.

In Exercises 1732, determine whether the graph of each equation is symmetric with respect to the y-axis, the x-axis, the origin, more than one of these, or none of these.

  1. 17. y=x2+6

  2. 18. y=x22

  3. 19. x=y2+6

  4. 20. x=y22

  5. 21. y2=x2+6

  6. 22. y2=x22

  7. 23. y=2x+3

  8. 24. y=2x+5

  9. 25. x2y3=2

  10. 26. x3y2=5

  11. 27. x2+y2=100

  12. 28. x2+y2=49

  13. 29. x2y2+3xy=1

  14. 30. x2y2+5xy=2

  15. 31. y4=x3+6

  16. 32. y5=x4+2

In Exercises 3336, use possible symmetry to determine whether each graph is the graph of an even function, an odd function, or a function that is neither even nor odd.

  1. 33.

    A graph is a curve with indefinite ends that passes through (negative 2, 4 fifths), (0, 4), and (2, 4 fifths).
  2. 34.

    A graph plots two curves with indefinite ends. The first curve passes through (1, 4), (2, 2), and (4, 1). The second curve passes through (negative 4, negative 1), (negative 2, negative 2), and (negative 1, negative 4).
  3. 35.

    A graph is a curve that passes through (negative 1, 1), (0, 0), and (1, negative 1).
  4. 36.

    A graph is a curve that passes through (negative 1, 3), (0, 2), (1, 1), and (2, negative 5).

In Exercises 3748, determine whether each function is even, odd, or neither. Then determine whether the function’s graph is symmetric with respect to the y-axis, the origin, or neither.

  1. 37. f(x)=x3+x

  2. 38. f(x)=x3x

  3. 39. g(x)=x2+x

  4. 40. g(x)=x2x

  5. 41. h(x)=x2x4

  6. 42. h(x)=2x2+x4

  7. 43. f(x)=x2x4+1

  8. 44. f(x)=2x2+x4+1

  9. 45. f(x)=15 x63x2

  10. 46. f(x)=2x36x5

  11. 47. f(x)=x1x2

  12. 48. f(x)=x21x2

  13. 49. Use the graph of f to determine each of the following. Where applicable, use interval notation.

    The graph of y = f of x is a horizontal line with an indefinite end from (negative 1, 4) to (0, 4) and a curve with an indefinite end that passes through (0, 4), (1, 0), (4, negative 4), and (7, 0).
    1. the domain of f

    2. the range of f

    3. the x-intercepts

    4. the y-intercept

    5. intervals on which f is increasing

    6. intervals on which f is decreasing

    7. intervals on which f is constant

    8. the number at which f has a relative minimum

    9. the relative minimum of f

    10. f(3)

    11. the values of x for which f(x)=2

    12. Is f even, odd, or neither?

  14. 50. Use the graph of f to determine each of the following. Where applicable, use interval notation.

    The graph of y = f of x is a curve with indefinite ends that passes through (negative 4, 0), (negative 2, 4), (0, 1), (3, 2), and (4, 0).
    1. the domain of f

    2. the range of f

    3. the x-intercepts

    4. the y-intercept

    5. intervals on which f is increasing

    6. intervals on which f is decreasing

    7. values of x for which f(x)0

    8. the numbers at which f has a relative maximum

    9. the relative maxima of f

    10. f(2)

    11. the values of x for which f(x)=0

    12. Is f even, odd, or neither?

  15. 51. Use the graph of f to determine each of the following. Where applicable, use interval notation.

    The graph of y = f of x is a curve with indefinite ends that starts at (3, 0), passes through (1, 4), (0, 3), and (negative 3, 0).
    1. the domain of f

    2. the range of f

    3. the zeros of f

    4. f(0)

    5. intervals on which f is increasing

    6. intervals on which f is decreasing

    7. values of x for which f(x)0

    8. any relative maxima and the numbers at which they occur

    9. the value of x for which f(x)=4

    10. Is f(1) positive or negative?

  16. 52. Use the graph of f to determine each of the following. Where applicable, use interval notation.

    The graph of y = f of x is a line that rises through (negative 5, negative 2) to (negative 2, 1), a horizontal line from (negative 2, 1) to (2, 1), and a line that falls to a closed point (5, negative 2).
    1. the domain of f

    2. the range of f

    3. the zeros of f

    4. f(0)

    5. intervals on which f is increasing

    6. intervals on which f is decreasing

    7. intervals on which f is constant

    8. values of x for which f(x)>0

    9. values of x for which f(x)=2

    10. Is f(4) positive or negative?

    11. Is f even, odd, or neither?

    12. Is f(2) a relative maximum?

In Exercises 5358, evaluate each piecewise function at the given values of the independent variable.

  1. 53. f(x)={3x+5ifx<04x+7ifx0

    1. f(2)

    2. f(0)

    3. f(3)

  2. 54. f(x)={6x1ifx<07x+3ifx0

    1. f(3)

    2. f(0)

    3. f(4)

  3. 55. g(x)={x+3ifx3(x+3)ifx<3

    1. g(0)

    2. g(6)

    3. g(3)

  4. 56. g(x)={x+5ifx5(x+5)ifx<5

    1. g(0)

    2. g(6)

    3. g(5)

  5. 57. h(x)={x29x3ifx36ifx=3

    1. h(5)

    2. h(0)

    3. h(3)

  6. 58. h(x)={x225x5ifx510ifx=5

    1. h(7)

    2. h(0)

    3. h(5)

In Exercises 5970, the domain of each piecewise function is (, ).

  1. Graph each function.

  2. Use your graph to determine the function’s range.

  1. 59. f(x)={xifx<0xifx0

  2. 60. f(x)={xifx<0xifx0

  3. 61. f(x)={2xifx02ifx>0

  4. 62. f(x)={12 xifx03ifx>0

  5. 63. f(x)={x+3ifx<2x3ifx2

  6. 64. f(x)={x+2ifx<3x2ifx3

  7. 65. f(x)={3ifx13ifx>1

  8. 66. f(x)={4ifx14ifx>1

  9. 67. f(x)={12 x2ifx<12x1ifx1

  10. 68. f(x)={ 12 x2ifx<12x+1ifx1

  11. 69. f(x)={0ifx<4xif4x<0x2ifx0

  12. 70. f(x)={0ifx<3 xif3x<0x21ifx0

In Exercises 7192, find and simplify the difference quotient

f(x+h)f(x)h, h0

for the given function.

  1. 71. f(x)=4x

  2. 72. f(x)=7x

  3. 73. f(x)=3x+7

  4. 74. f(x)=6x+1

  5. 75. f(x)=x2

  6. 76. f(x)=2x2

  7. 77. f(x)=x24x+3

  8. 78. f(x)=x25x+8

  9. 79. f(x)=2x2+x1

  10. 80. f(x)=3x2+x+5

  11. 81. f(x)=x2+2x+4

  12. 82. f(x)=x23x+1

  13. 83. f(x)=2x2+5x+7

  14. 84. f(x)=3x2+2x1

  15. 85. f(x)=2x2x+3

  16. 86. f(x)=3x2+x1

  17. 87. f(x)=6

  18. 88. f(x)=7

  19. 89. f(x)=1x

  20. 90. f(x)=12x

  21. 91. f(x)=x

  22. 92. f(x)=x1

Practice PLUS

In Exercises 9394, let f be defined by the following graph:

The image shows a stepwise function graph.
  1. 93. Find

    f(1.5)+f(0.9)[f(π)]2+f(3)÷f(1)f(π).
  2. 94. Find

    f(2.5)f(1.9)[f(π)]2+f(3)÷f(1)f(π).

A cable company offers the following high-speed Internet plans. Also given are the piecewise functions that model these plans. Use this information to solve Exercises 9596.

Plan A

  • $40 per month includes 400 GB of data.

  • Additional data costs $0.20 per GB.

    C(g)={40if0g40040+0.20(g400)ifg>400

Plan B

  • $60 per month includes 1000 GB of data.

  • Additional data costs $0.20 per GB.

    C(g)={60if0g100060+0.20(g1000)ifg>1000
  1. 95. Simplify the algebraic expression in the second line of the piecewise function for plan A. Then use point-plotting to graph the function.

  2. 96. Simplify the algebraic expression in the second line of the piecewise function for plan B. Then use point-plotting to graph the function.

In Exercises 9798, write a piecewise function that models each high-speed Internet plan. Then graph the function.

  1. 97. $50 per month includes 600 GB. Additional data costs $0.30 per GB.

  2. 98. $80 per month includes 2000 GB. Additional data costs $0.35 per GB.

Application Exercises

With aging, body fat increases and muscle mass declines. The line graphs show the percent body fat in adult women and men as they age from 25 to 75 years. Use the graphs to solve Exercises 99106.

A graph plots the percent of body fat in adults.

Source: Thompson et al., The Science of Nutrition, Benjamin Cummings, 2008

  1. 99. State the intervals on which the graph giving the percent body fat in women is increasing and decreasing.

  2. 100. State the intervals on which the graph giving the percent body fat in men is increasing and decreasing.

  3. 101. For what age does the percent body fat in women reach a maximum? What is the percent body fat for that age?

  4. 102. At what age does the percent body fat in men reach a maximum? What is the percent body fat for that age?

  5. 103. Use interval notation to give the domain and the range for the graph of the function for women.

  6. 104. Use interval notation to give the domain and the range for the graph of the function for men.

  7. 105. The function p(x)=0.002x2+0.15x+22.86 models percent body fat, p(x), where x is the number of years a person’s age exceeds 25. Use the graphs to determine whether this model describes percent body fat in women or in men.

  8. 106. The function p(x)=0.004x2+0.25x+33.64 models percent body fat, p(x), where x is the number of years a person’s age exceeds 25. Use the graphs to determine whether this model describes percent body fat in women or in men.

Here is the Federal Tax Rate Schedule X that specifies the tax owed by a single taxpayer for 2020.

If Your Taxable Income Is Over But Not Over The Tax You Owe Is Of the Amount Over
$          0 $    9,875                       10% $          0
$    9875 $  40,125 $   987.50+12% $    9875
$  40,125 $  85,525 $   4617.50+22% $  40,125
$  85,525 $163,300 $14,605.50+24% $  85,525
$163,300 $207,350 $33,271.50+32% $163,300
$207,350 $518,400 $47,367.50+35% $207,350
$518,400       − $   156,235+37% $518,400

The preceding tax table can be modeled by a piecewise function, where x represents the taxable income of a single taxpayer and T(x) is the tax owed:

T(x)={0.10xif0<x9875987.50+0.12(x9875)if9875<x40,1254617.50+0.22(x40,125)if40,125<x85,52514,605.50+0.24(x85,525)if85,525<x163,30033,271.50+0.32(x163,300)if163,300<x207,350wewdwdwdss?sdwdwdwddw¯if207,350<x518,400sssasassxxax?xssceddvdddd¯ifx>518,400

Use this information to solve Exercises 107108.

  1. 107. Find and interpret T(35,000).

  2. 108. Find and interpret T(50,000).

In Exercises 109110, refer to the preceding tax table.

  1. 109. Find the algebraic expression for the missing piece of T(x) that models tax owed for the domain (207,350,518,400].

  2. 110. Find the algebraic expression for the missing piece of T(x) that models tax owed for the domain (518,400, ).

The figure shows the cost of mailing a first-class letter, f(x), as a function of its weight, x, in ounces, in June 2020. Use the graph to solve Exercises 111114.

The image shows a function graph that plots cost, in dollars, versus weight, in ounces.

Source: Lynn E. Baring, Postmaster, Inverness, CA

  1. 111. Find f(3). What does this mean in terms of the variables in this situation?

  2. 112. Find f(3.75). What does this mean in terms of the variables in this situation?

  3. 113. What is the cost of mailing a letter that weighs 1.5 ounces?

  4. 114. What is the cost of mailing a letter that weighs 1.8 ounces?

  5. 115. Furry Finances. A pet insurance policy has a monthly rate that is a function of the age of the insured dog or cat. For pets whose age does not exceed 4, the monthly cost is $20. The cost then increases by $2 for each successive year of the pet’s age.

    Age Not Exceeding Monthly Cost
    4 $20
    5 $22
    6 $24

    The cost schedule continues in this manner for ages not exceeding 10. The cost for pets whose ages exceed 10 is $40. Use this information to create a graph that shows the monthly cost of the insurance, f(x), for a pet of age x, where the function’s domain is [0, 14].

Explaining the Concepts

  1. 116. What does it mean if a function f is increasing on an interval?

  2. 117. Suppose that a function f whose graph contains no breaks or gaps on (a, c) is increasing on (a, b), decreasing on (b, c), and defined at b. Describe what occurs at x=b. What does the function value f(b) represent?

  3. 118. Given an equation in x and y, how do you determine if its graph is symmetric with respect to the y-axis?

  4. 119. Given an equation in x and y, how do you determine if its graph is symmetric with respect to the x-axis?

  5. 120. Given an equation in x and y, how do you determine if its graph is symmetric with respect to the origin?

  6. 121. If you are given a function’s graph, how do you determine if the function is even, odd, or neither?

  7. 122. If you are given a function’s equation, how do you determine if the function is even, odd, or neither?

  8. 123. What is a piecewise function?

  9. 124. Explain how to find the difference quotient of a function f, f(x+h)f(x)h, if an equation for f is given.

Technology Exercises

  1. 125. The function

    f(x)=0.00002x3+0.008x20.3x+6.95

    models the number of annual physician visits, f(x), by a person of age x. Graph the function in a [0,100,5] by [0,40,2] viewing rectangle. What does the shape of the graph indicate about the relationship between one’s age and the number of annual physician visits? Use the  TABLE  or minimum function capability to find the coordinates of the minimum point on the graph of the function. What does this mean?

In Exercises 126131, use a graphing utility to graph each function. Use a [5, 5, 1] by [5, 5, 1] viewing rectangle. Then find the intervals on which the function is increasing, decreasing, or constant.

  1. 126. f(x)=x36x2+9x+1

  2. 127. g(x)=|4x2|

  3. 128. h(x)=|x2|+|x+2|

  4. 129. f(x)=x13(x4)

  5. 130. g(x)=x23

  6. 131. h(x)=2x25

  7. 132.

    1. Graph the functions f(x)=xn for n=2, 4, and 6 in a [2, 2, 1] by [1, 3, 1] viewing rectangle.

    2. Graph the functions f(x)=xn for n=1, 3, and 5 in a [2, 2, 1] by [2, 2, 1] viewing rectangle.

    3. If n is positive and even, where is the graph of f(x)=xn increasing and where is it decreasing?

    4. If n is positive and odd, what can you conclude about the graph of f(x)=xn in terms of increasing or decreasing behavior?

    5. Graph all six functions in a [1, 3, 1] by [1, 3, 1] viewing rectangle. What do you observe about the graphs in terms of how flat or how steep they are?

Critical Thinking Exercises

Make Sense? In Exercises 133136, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 133. My graph is decreasing on (, a) and increasing on (a, ), so f(a) must be a relative maximum.

  2. 134. This work by artist Scott Kim has the same kind of symmetry as an even function.

    The word Dyslexia is written such that the alphabets on either side of the letter, l, appear as a mirror image of each other. Double prime DYSLEXIA comma double prime 19 81.
  3. 135. I graphed

    f(x)={2ifx43ifx=4

    and one piece of my graph is a single point.

  4. 136. I noticed that the difference quotient is always zero if f(x)=c, where c is any constant.

  5. 137. Sketch the graph of f using the following properties. (More than one correct graph is possible.) f is a piecewise function that is decreasing on (, 2), f(2)=0, f is increasing on (2, ), and the range of f is [0, ).

  6. 138. Define a piecewise function on the intervals (, 2], (2, 5), and [5, ) that does not “jump” at 2 or 5 such that one piece is a constant function, another piece is an increasing function, and the third piece is a decreasing function.

  7. 139. Suppose that h(x)=f(x)g(x). The function f can be even, odd, or neither. The same is true for the function g.

    1. Under what conditions is h definitely an even function?

    2. Under what conditions is h definitely an odd function?

Retaining the Concepts

  1. 140. You invested $80,000 in two accounts paying 1.5% and 1.7% annual interest. If the total interest earned for the year was $1320, how much was invested at each rate? (Section P.8, Example 5)

  2. 141. Solve for A: C=A+Ar.

    (Section P.7, Example 5)

  3. 142. Solve by the quadratic formula: 5x26x8=0.

    (Section P.7, Example 10)

Preview Exercises

Exercises 143145 will help you prepare for the material covered in the next section.

  1. 143. If (x1, y1)=(3, 1) and (x2, y2)=(2, 4), find y2y1x2x1.

  2. 144. Find the ordered pairs (_______, 0) and (0, _______) satisfying 4x3y6=0.

  3. 145. Solve for y: 3x+2y4=0.

1.3: Exercise Set

1.3 Exercise Set

Practice Exercises

In Exercises 112, use the graph to determine

  1. intervals on which the function is increasing, if any.

  2. intervals on which the function is decreasing, if any.

  3. intervals on which the function is constant, if any.

  1. 1.

    A graph is an upward opening parabola with indefinite ends that falls from (negative 4, 5) through (negative 3, 0) to its vertex at (negative 1, negative 4), and then rises through (1, 0) to point (4, 4).
  2. 2.

    A graph is a downward opening parabola with indefinite ends that rises from (negative 4, negative 5) through (negative 3, 0) to its vertex at (negative 1, 4) and falls through (1, 0) to point (2, negative 5). All values are estimated.
  3. 3.

    A graph is a curve with a closed point and indefinite end at (0, 1) and passes through (4, 3). All values are estimated.
  4. 4.

    A graph is a curve with a closed point and indefinite end that starts at (negative 1, 0) and passes through (0, 1) and (3, 2). All values are estimated.
  5. 5.

    A graph is a line with closed points that falls from (negative 2, 6), passes through (0, 4), (4, 0), (0, 4) and ends at (6, negative 2). All values are estimated.
  6. 6.

    A graph is a line with closed points that rises from (negative 3, negative 5) (negative 0.5, 0), passes through (0, 1), and ends at (2, 5). All values are estimated.
  7. 7.

    A graph is a line that rises through (negative 4, negative 5) to (negative 1, 2) and then extends horizontally through (0, negative 2) and (4, negative 2). All values are estimated.
  8. 8.

    A graph is a line that passes along the negative x axis to the origin and then rises through (1, 2) and (2, 4). All values are estimated.
  9. 9.

    A graph is an m shaped curve with indefinite ends that start in the third quadrant and end in the fourth quadrant.
  10. 10.

    A graph is a sinusoidal curve with closed ends that starts at (negative 5, 0), passes through (negative 4, 4), (negative 3, 0), (negative 2, negative 4), (negative 1, 0), (0, 4), (1, 0), (2, negative 4), (3, 0), (4, 4), and ends at (5, 0).
  11. 11.

    A graph is a line curve with indefinite ends that passes through (negative 3, negative 2), (negative 2, negative 2), (0, negative 1), (2, 0), (4, 2), and (5, 2).
  12. 12.

    A graph is a curve with indefinite ends that passes through (negative 5, 2), (negative 4, 2), (0, 0), (2, negative 2), and (3, negative 2).

In Exercises 1316, the graph of a function f is given. Use the graph to find each of the following:

  1. The numbers, if any, at which f has a relative maximum. What are these relative maxima?

  2. The numbers, if any, at which f has a relative minimum. What are these relative minima?

  1. 13.

    A graph is a w shaped curve with indefinite ends that passes through (negative 4, 3), (negative 3, 0), (0, 4), (3, 0), and (4, 3).
  2. 14.

    A graph is a curve that passes through (negative 3, negative 1), (0, 2), and (3, negative 1).
  3. 15.

    A graphing calculator screen displays a N shaped curve, f of x = 2 x cubed + 3 x squared minus 12 x + 1, which passes through the third quadrant, second quadrant, fourth quadrant, and then first quadrant.

  4. 16.

    A graphing calculator screen displays a curve, f of x = 2 x cubed minus 15 x squared + 24 x + 19, which passes through the third, second, and first quadrant.

In Exercises 1732, determine whether the graph of each equation is symmetric with respect to the y-axis, the x-axis, the origin, more than one of these, or none of these.

  1. 17. y=x2+6

  2. 18. y=x22

  3. 19. x=y2+6

  4. 20. x=y22

  5. 21. y2=x2+6

  6. 22. y2=x22

  7. 23. y=2x+3

  8. 24. y=2x+5

  9. 25. x2y3=2

  10. 26. x3y2=5

  11. 27. x2+y2=100

  12. 28. x2+y2=49

  13. 29. x2y2+3xy=1

  14. 30. x2y2+5xy=2

  15. 31. y4=x3+6

  16. 32. y5=x4+2

In Exercises 3336, use possible symmetry to determine whether each graph is the graph of an even function, an odd function, or a function that is neither even nor odd.

  1. 33.

    A graph is a curve with indefinite ends that passes through (negative 2, 4 fifths), (0, 4), and (2, 4 fifths).
  2. 34.

    A graph plots two curves with indefinite ends. The first curve passes through (1, 4), (2, 2), and (4, 1). The second curve passes through (negative 4, negative 1), (negative 2, negative 2), and (negative 1, negative 4).
  3. 35.

    A graph is a curve that passes through (negative 1, 1), (0, 0), and (1, negative 1).
  4. 36.

    A graph is a curve that passes through (negative 1, 3), (0, 2), (1, 1), and (2, negative 5).

In Exercises 3748, determine whether each function is even, odd, or neither. Then determine whether the function’s graph is symmetric with respect to the y-axis, the origin, or neither.

  1. 37. f(x)=x3+x

  2. 38. f(x)=x3x

  3. 39. g(x)=x2+x

  4. 40. g(x)=x2x

  5. 41. h(x)=x2x4

  6. 42. h(x)=2x2+x4

  7. 43. f(x)=x2x4+1

  8. 44. f(x)=2x2+x4+1

  9. 45. f(x)=15 x63x2

  10. 46. f(x)=2x36x5

  11. 47. f(x)=x1x2

  12. 48. f(x)=x21x2

  13. 49. Use the graph of f to determine each of the following. Where applicable, use interval notation.

    The graph of y = f of x is a horizontal line with an indefinite end from (negative 1, 4) to (0, 4) and a curve with an indefinite end that passes through (0, 4), (1, 0), (4, negative 4), and (7, 0).
    1. the domain of f

    2. the range of f

    3. the x-intercepts

    4. the y-intercept

    5. intervals on which f is increasing

    6. intervals on which f is decreasing

    7. intervals on which f is constant

    8. the number at which f has a relative minimum

    9. the relative minimum of f

    10. f(3)

    11. the values of x for which f(x)=2

    12. Is f even, odd, or neither?

  14. 50. Use the graph of f to determine each of the following. Where applicable, use interval notation.

    The graph of y = f of x is a curve with indefinite ends that passes through (negative 4, 0), (negative 2, 4), (0, 1), (3, 2), and (4, 0).
    1. the domain of f

    2. the range of f

    3. the x-intercepts

    4. the y-intercept

    5. intervals on which f is increasing

    6. intervals on which f is decreasing

    7. values of x for which f(x)0

    8. the numbers at which f has a relative maximum

    9. the relative maxima of f

    10. f(2)

    11. the values of x for which f(x)=0

    12. Is f even, odd, or neither?

  15. 51. Use the graph of f to determine each of the following. Where applicable, use interval notation.

    The graph of y = f of x is a curve with indefinite ends that starts at (3, 0), passes through (1, 4), (0, 3), and (negative 3, 0).
    1. the domain of f

    2. the range of f

    3. the zeros of f

    4. f(0)

    5. intervals on which f is increasing

    6. intervals on which f is decreasing

    7. values of x for which f(x)0

    8. any relative maxima and the numbers at which they occur

    9. the value of x for which f(x)=4

    10. Is f(1) positive or negative?

  16. 52. Use the graph of f to determine each of the following. Where applicable, use interval notation.

    The graph of y = f of x is a line that rises through (negative 5, negative 2) to (negative 2, 1), a horizontal line from (negative 2, 1) to (2, 1), and a line that falls to a closed point (5, negative 2).
    1. the domain of f

    2. the range of f

    3. the zeros of f

    4. f(0)

    5. intervals on which f is increasing

    6. intervals on which f is decreasing

    7. intervals on which f is constant

    8. values of x for which f(x)>0

    9. values of x for which f(x)=2

    10. Is f(4) positive or negative?

    11. Is f even, odd, or neither?

    12. Is f(2) a relative maximum?

In Exercises 5358, evaluate each piecewise function at the given values of the independent variable.

  1. 53. f(x)={3x+5ifx<04x+7ifx0

    1. f(2)

    2. f(0)

    3. f(3)

  2. 54. f(x)={6x1ifx<07x+3ifx0

    1. f(3)

    2. f(0)

    3. f(4)

  3. 55. g(x)={x+3ifx3(x+3)ifx<3

    1. g(0)

    2. g(6)

    3. g(3)

  4. 56. g(x)={x+5ifx5(x+5)ifx<5

    1. g(0)

    2. g(6)

    3. g(5)

  5. 57. h(x)={x29x3ifx36ifx=3

    1. h(5)

    2. h(0)

    3. h(3)

  6. 58. h(x)={x225x5ifx510ifx=5

    1. h(7)

    2. h(0)

    3. h(5)

In Exercises 5970, the domain of each piecewise function is (, ).

  1. Graph each function.

  2. Use your graph to determine the function’s range.

  1. 59. f(x)={xifx<0xifx0

  2. 60. f(x)={xifx<0xifx0

  3. 61. f(x)={2xifx02ifx>0

  4. 62. f(x)={12 xifx03ifx>0

  5. 63. f(x)={x+3ifx<2x3ifx2

  6. 64. f(x)={x+2ifx<3x2ifx3

  7. 65. f(x)={3ifx13ifx>1

  8. 66. f(x)={4ifx14ifx>1

  9. 67. f(x)={12 x2ifx<12x1ifx1

  10. 68. f(x)={ 12 x2ifx<12x+1ifx1

  11. 69. f(x)={0ifx<4xif4x<0x2ifx0

  12. 70. f(x)={0ifx<3 xif3x<0x21ifx0

In Exercises 7192, find and simplify the difference quotient

f(x+h)f(x)h, h0

for the given function.

  1. 71. f(x)=4x

  2. 72. f(x)=7x

  3. 73. f(x)=3x+7

  4. 74. f(x)=6x+1

  5. 75. f(x)=x2

  6. 76. f(x)=2x2

  7. 77. f(x)=x24x+3

  8. 78. f(x)=x25x+8

  9. 79. f(x)=2x2+x1

  10. 80. f(x)=3x2+x+5

  11. 81. f(x)=x2+2x+4

  12. 82. f(x)=x23x+1

  13. 83. f(x)=2x2+5x+7

  14. 84. f(x)=3x2+2x1

  15. 85. f(x)=2x2x+3

  16. 86. f(x)=3x2+x1

  17. 87. f(x)=6

  18. 88. f(x)=7

  19. 89. f(x)=1x

  20. 90. f(x)=12x

  21. 91. f(x)=x

  22. 92. f(x)=x1

Practice PLUS

In Exercises 9394, let f be defined by the following graph:

The image shows a stepwise function graph.
  1. 93. Find

    f(1.5)+f(0.9)[f(π)]2+f(3)÷f(1)f(π).
  2. 94. Find

    f(2.5)f(1.9)[f(π)]2+f(3)÷f(1)f(π).

A cable company offers the following high-speed Internet plans. Also given are the piecewise functions that model these plans. Use this information to solve Exercises 9596.

Plan A

  • $40 per month includes 400 GB of data.

  • Additional data costs $0.20 per GB.

    C(g)={40if0g40040+0.20(g400)ifg>400

Plan B

  • $60 per month includes 1000 GB of data.

  • Additional data costs $0.20 per GB.

    C(g)={60if0g100060+0.20(g1000)ifg>1000
  1. 95. Simplify the algebraic expression in the second line of the piecewise function for plan A. Then use point-plotting to graph the function.

  2. 96. Simplify the algebraic expression in the second line of the piecewise function for plan B. Then use point-plotting to graph the function.

In Exercises 9798, write a piecewise function that models each high-speed Internet plan. Then graph the function.

  1. 97. $50 per month includes 600 GB. Additional data costs $0.30 per GB.

  2. 98. $80 per month includes 2000 GB. Additional data costs $0.35 per GB.

Application Exercises

With aging, body fat increases and muscle mass declines. The line graphs show the percent body fat in adult women and men as they age from 25 to 75 years. Use the graphs to solve Exercises 99106.

A graph plots the percent of body fat in adults.

Source: Thompson et al., The Science of Nutrition, Benjamin Cummings, 2008

  1. 99. State the intervals on which the graph giving the percent body fat in women is increasing and decreasing.

  2. 100. State the intervals on which the graph giving the percent body fat in men is increasing and decreasing.

  3. 101. For what age does the percent body fat in women reach a maximum? What is the percent body fat for that age?

  4. 102. At what age does the percent body fat in men reach a maximum? What is the percent body fat for that age?

  5. 103. Use interval notation to give the domain and the range for the graph of the function for women.

  6. 104. Use interval notation to give the domain and the range for the graph of the function for men.

  7. 105. The function p(x)=0.002x2+0.15x+22.86 models percent body fat, p(x), where x is the number of years a person’s age exceeds 25. Use the graphs to determine whether this model describes percent body fat in women or in men.

  8. 106. The function p(x)=0.004x2+0.25x+33.64 models percent body fat, p(x), where x is the number of years a person’s age exceeds 25. Use the graphs to determine whether this model describes percent body fat in women or in men.

Here is the Federal Tax Rate Schedule X that specifies the tax owed by a single taxpayer for 2020.

If Your Taxable Income Is Over But Not Over The Tax You Owe Is Of the Amount Over
$          0 $    9,875                       10% $          0
$    9875 $  40,125 $   987.50+12% $    9875
$  40,125 $  85,525 $   4617.50+22% $  40,125
$  85,525 $163,300 $14,605.50+24% $  85,525
$163,300 $207,350 $33,271.50+32% $163,300
$207,350 $518,400 $47,367.50+35% $207,350
$518,400       − $   156,235+37% $518,400

The preceding tax table can be modeled by a piecewise function, where x represents the taxable income of a single taxpayer and T(x) is the tax owed:

T(x)={0.10xif0<x9875987.50+0.12(x9875)if9875<x40,1254617.50+0.22(x40,125)if40,125<x85,52514,605.50+0.24(x85,525)if85,525<x163,30033,271.50+0.32(x163,300)if163,300<x207,350wewdwdwdss?sdwdwdwddw¯if207,350<x518,400sssasassxxax?xssceddvdddd¯ifx>518,400

Use this information to solve Exercises 107108.

  1. 107. Find and interpret T(35,000).

  2. 108. Find and interpret T(50,000).

In Exercises 109110, refer to the preceding tax table.

  1. 109. Find the algebraic expression for the missing piece of T(x) that models tax owed for the domain (207,350,518,400].

  2. 110. Find the algebraic expression for the missing piece of T(x) that models tax owed for the domain (518,400, ).

The figure shows the cost of mailing a first-class letter, f(x), as a function of its weight, x, in ounces, in June 2020. Use the graph to solve Exercises 111114.

The image shows a function graph that plots cost, in dollars, versus weight, in ounces.

Source: Lynn E. Baring, Postmaster, Inverness, CA

  1. 111. Find f(3). What does this mean in terms of the variables in this situation?

  2. 112. Find f(3.75). What does this mean in terms of the variables in this situation?

  3. 113. What is the cost of mailing a letter that weighs 1.5 ounces?

  4. 114. What is the cost of mailing a letter that weighs 1.8 ounces?

  5. 115. Furry Finances. A pet insurance policy has a monthly rate that is a function of the age of the insured dog or cat. For pets whose age does not exceed 4, the monthly cost is $20. The cost then increases by $2 for each successive year of the pet’s age.

    Age Not Exceeding Monthly Cost
    4 $20
    5 $22
    6 $24

    The cost schedule continues in this manner for ages not exceeding 10. The cost for pets whose ages exceed 10 is $40. Use this information to create a graph that shows the monthly cost of the insurance, f(x), for a pet of age x, where the function’s domain is [0, 14].

Explaining the Concepts

  1. 116. What does it mean if a function f is increasing on an interval?

  2. 117. Suppose that a function f whose graph contains no breaks or gaps on (a, c) is increasing on (a, b), decreasing on (b, c), and defined at b. Describe what occurs at x=b. What does the function value f(b) represent?

  3. 118. Given an equation in x and y, how do you determine if its graph is symmetric with respect to the y-axis?

  4. 119. Given an equation in x and y, how do you determine if its graph is symmetric with respect to the x-axis?

  5. 120. Given an equation in x and y, how do you determine if its graph is symmetric with respect to the origin?

  6. 121. If you are given a function’s graph, how do you determine if the function is even, odd, or neither?

  7. 122. If you are given a function’s equation, how do you determine if the function is even, odd, or neither?

  8. 123. What is a piecewise function?

  9. 124. Explain how to find the difference quotient of a function f, f(x+h)f(x)h, if an equation for f is given.

Technology Exercises

  1. 125. The function

    f(x)=0.00002x3+0.008x20.3x+6.95

    models the number of annual physician visits, f(x), by a person of age x. Graph the function in a [0,100,5] by [0,40,2] viewing rectangle. What does the shape of the graph indicate about the relationship between one’s age and the number of annual physician visits? Use the  TABLE  or minimum function capability to find the coordinates of the minimum point on the graph of the function. What does this mean?

In Exercises 126131, use a graphing utility to graph each function. Use a [5, 5, 1] by [5, 5, 1] viewing rectangle. Then find the intervals on which the function is increasing, decreasing, or constant.

  1. 126. f(x)=x36x2+9x+1

  2. 127. g(x)=|4x2|

  3. 128. h(x)=|x2|+|x+2|

  4. 129. f(x)=x13(x4)

  5. 130. g(x)=x23

  6. 131. h(x)=2x25

  7. 132.

    1. Graph the functions f(x)=xn for n=2, 4, and 6 in a [2, 2, 1] by [1, 3, 1] viewing rectangle.

    2. Graph the functions f(x)=xn for n=1, 3, and 5 in a [2, 2, 1] by [2, 2, 1] viewing rectangle.

    3. If n is positive and even, where is the graph of f(x)=xn increasing and where is it decreasing?

    4. If n is positive and odd, what can you conclude about the graph of f(x)=xn in terms of increasing or decreasing behavior?

    5. Graph all six functions in a [1, 3, 1] by [1, 3, 1] viewing rectangle. What do you observe about the graphs in terms of how flat or how steep they are?

Critical Thinking Exercises

Make Sense? In Exercises 133136, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 133. My graph is decreasing on (, a) and increasing on (a, ), so f(a) must be a relative maximum.

  2. 134. This work by artist Scott Kim has the same kind of symmetry as an even function.

    The word Dyslexia is written such that the alphabets on either side of the letter, l, appear as a mirror image of each other. Double prime DYSLEXIA comma double prime 19 81.
  3. 135. I graphed

    f(x)={2ifx43ifx=4

    and one piece of my graph is a single point.

  4. 136. I noticed that the difference quotient is always zero if f(x)=c, where c is any constant.

  5. 137. Sketch the graph of f using the following properties. (More than one correct graph is possible.) f is a piecewise function that is decreasing on (, 2), f(2)=0, f is increasing on (2, ), and the range of f is [0, ).

  6. 138. Define a piecewise function on the intervals (, 2], (2, 5), and [5, ) that does not “jump” at 2 or 5 such that one piece is a constant function, another piece is an increasing function, and the third piece is a decreasing function.

  7. 139. Suppose that h(x)=f(x)g(x). The function f can be even, odd, or neither. The same is true for the function g.

    1. Under what conditions is h definitely an even function?

    2. Under what conditions is h definitely an odd function?

Retaining the Concepts

  1. 140. You invested $80,000 in two accounts paying 1.5% and 1.7% annual interest. If the total interest earned for the year was $1320, how much was invested at each rate? (Section P.8, Example 5)

  2. 141. Solve for A: C=A+Ar.

    (Section P.7, Example 5)

  3. 142. Solve by the quadratic formula: 5x26x8=0.

    (Section P.7, Example 10)

Preview Exercises

Exercises 143145 will help you prepare for the material covered in the next section.

  1. 143. If (x1, y1)=(3, 1) and (x2, y2)=(2, 4), find y2y1x2x1.

  2. 144. Find the ordered pairs (_______, 0) and (0, _______) satisfying 4x3y6=0.

  3. 145. Solve for y: 3x+2y4=0.

1.3: Exercise Set

1.3 Exercise Set

Practice Exercises

In Exercises 112, use the graph to determine

  1. intervals on which the function is increasing, if any.

  2. intervals on which the function is decreasing, if any.

  3. intervals on which the function is constant, if any.

  1. 1.

    A graph is an upward opening parabola with indefinite ends that falls from (negative 4, 5) through (negative 3, 0) to its vertex at (negative 1, negative 4), and then rises through (1, 0) to point (4, 4).
  2. 2.

    A graph is a downward opening parabola with indefinite ends that rises from (negative 4, negative 5) through (negative 3, 0) to its vertex at (negative 1, 4) and falls through (1, 0) to point (2, negative 5). All values are estimated.
  3. 3.

    A graph is a curve with a closed point and indefinite end at (0, 1) and passes through (4, 3). All values are estimated.
  4. 4.

    A graph is a curve with a closed point and indefinite end that starts at (negative 1, 0) and passes through (0, 1) and (3, 2). All values are estimated.
  5. 5.

    A graph is a line with closed points that falls from (negative 2, 6), passes through (0, 4), (4, 0), (0, 4) and ends at (6, negative 2). All values are estimated.
  6. 6.

    A graph is a line with closed points that rises from (negative 3, negative 5) (negative 0.5, 0), passes through (0, 1), and ends at (2, 5). All values are estimated.
  7. 7.

    A graph is a line that rises through (negative 4, negative 5) to (negative 1, 2) and then extends horizontally through (0, negative 2) and (4, negative 2). All values are estimated.
  8. 8.

    A graph is a line that passes along the negative x axis to the origin and then rises through (1, 2) and (2, 4). All values are estimated.
  9. 9.

    A graph is an m shaped curve with indefinite ends that start in the third quadrant and end in the fourth quadrant.
  10. 10.

    A graph is a sinusoidal curve with closed ends that starts at (negative 5, 0), passes through (negative 4, 4), (negative 3, 0), (negative 2, negative 4), (negative 1, 0), (0, 4), (1, 0), (2, negative 4), (3, 0), (4, 4), and ends at (5, 0).
  11. 11.

    A graph is a line curve with indefinite ends that passes through (negative 3, negative 2), (negative 2, negative 2), (0, negative 1), (2, 0), (4, 2), and (5, 2).
  12. 12.

    A graph is a curve with indefinite ends that passes through (negative 5, 2), (negative 4, 2), (0, 0), (2, negative 2), and (3, negative 2).

In Exercises 1316, the graph of a function f is given. Use the graph to find each of the following:

  1. The numbers, if any, at which f has a relative maximum. What are these relative maxima?

  2. The numbers, if any, at which f has a relative minimum. What are these relative minima?

  1. 13.

    A graph is a w shaped curve with indefinite ends that passes through (negative 4, 3), (negative 3, 0), (0, 4), (3, 0), and (4, 3).
  2. 14.

    A graph is a curve that passes through (negative 3, negative 1), (0, 2), and (3, negative 1).
  3. 15.

    A graphing calculator screen displays a N shaped curve, f of x = 2 x cubed + 3 x squared minus 12 x + 1, which passes through the third quadrant, second quadrant, fourth quadrant, and then first quadrant.

  4. 16.

    A graphing calculator screen displays a curve, f of x = 2 x cubed minus 15 x squared + 24 x + 19, which passes through the third, second, and first quadrant.

In Exercises 1732, determine whether the graph of each equation is symmetric with respect to the y-axis, the x-axis, the origin, more than one of these, or none of these.

  1. 17. y=x2+6

  2. 18. y=x22

  3. 19. x=y2+6

  4. 20. x=y22

  5. 21. y2=x2+6

  6. 22. y2=x22

  7. 23. y=2x+3

  8. 24. y=2x+5

  9. 25. x2y3=2

  10. 26. x3y2=5

  11. 27. x2+y2=100

  12. 28. x2+y2=49

  13. 29. x2y2+3xy=1

  14. 30. x2y2+5xy=2

  15. 31. y4=x3+6

  16. 32. y5=x4+2

In Exercises 3336, use possible symmetry to determine whether each graph is the graph of an even function, an odd function, or a function that is neither even nor odd.

  1. 33.

    A graph is a curve with indefinite ends that passes through (negative 2, 4 fifths), (0, 4), and (2, 4 fifths).
  2. 34.

    A graph plots two curves with indefinite ends. The first curve passes through (1, 4), (2, 2), and (4, 1). The second curve passes through (negative 4, negative 1), (negative 2, negative 2), and (negative 1, negative 4).
  3. 35.

    A graph is a curve that passes through (negative 1, 1), (0, 0), and (1, negative 1).
  4. 36.

    A graph is a curve that passes through (negative 1, 3), (0, 2), (1, 1), and (2, negative 5).

In Exercises 3748, determine whether each function is even, odd, or neither. Then determine whether the function’s graph is symmetric with respect to the y-axis, the origin, or neither.

  1. 37. f(x)=x3+x

  2. 38. f(x)=x3x

  3. 39. g(x)=x2+x

  4. 40. g(x)=x2x

  5. 41. h(x)=x2x4

  6. 42. h(x)=2x2+x4

  7. 43. f(x)=x2x4+1

  8. 44. f(x)=2x2+x4+1

  9. 45. f(x)=15 x63x2

  10. 46. f(x)=2x36x5

  11. 47. f(x)=x1x2

  12. 48. f(x)=x21x2

  13. 49. Use the graph of f to determine each of the following. Where applicable, use interval notation.

    The graph of y = f of x is a horizontal line with an indefinite end from (negative 1, 4) to (0, 4) and a curve with an indefinite end that passes through (0, 4), (1, 0), (4, negative 4), and (7, 0).
    1. the domain of f

    2. the range of f

    3. the x-intercepts

    4. the y-intercept

    5. intervals on which f is increasing

    6. intervals on which f is decreasing

    7. intervals on which f is constant

    8. the number at which f has a relative minimum

    9. the relative minimum of f

    10. f(3)

    11. the values of x for which f(x)=2

    12. Is f even, odd, or neither?

  14. 50. Use the graph of f to determine each of the following. Where applicable, use interval notation.

    The graph of y = f of x is a curve with indefinite ends that passes through (negative 4, 0), (negative 2, 4), (0, 1), (3, 2), and (4, 0).
    1. the domain of f

    2. the range of f

    3. the x-intercepts

    4. the y-intercept

    5. intervals on which f is increasing

    6. intervals on which f is decreasing

    7. values of x for which f(x)0

    8. the numbers at which f has a relative maximum

    9. the relative maxima of f

    10. f(2)

    11. the values of x for which f(x)=0

    12. Is f even, odd, or neither?

  15. 51. Use the graph of f to determine each of the following. Where applicable, use interval notation.

    The graph of y = f of x is a curve with indefinite ends that starts at (3, 0), passes through (1, 4), (0, 3), and (negative 3, 0).
    1. the domain of f

    2. the range of f

    3. the zeros of f

    4. f(0)

    5. intervals on which f is increasing

    6. intervals on which f is decreasing

    7. values of x for which f(x)0

    8. any relative maxima and the numbers at which they occur

    9. the value of x for which f(x)=4

    10. Is f(1) positive or negative?

  16. 52. Use the graph of f to determine each of the following. Where applicable, use interval notation.

    The graph of y = f of x is a line that rises through (negative 5, negative 2) to (negative 2, 1), a horizontal line from (negative 2, 1) to (2, 1), and a line that falls to a closed point (5, negative 2).
    1. the domain of f

    2. the range of f

    3. the zeros of f

    4. f(0)

    5. intervals on which f is increasing

    6. intervals on which f is decreasing

    7. intervals on which f is constant

    8. values of x for which f(x)>0

    9. values of x for which f(x)=2

    10. Is f(4) positive or negative?

    11. Is f even, odd, or neither?

    12. Is f(2) a relative maximum?

In Exercises 5358, evaluate each piecewise function at the given values of the independent variable.

  1. 53. f(x)={3x+5ifx<04x+7ifx0

    1. f(2)

    2. f(0)

    3. f(3)

  2. 54. f(x)={6x1ifx<07x+3ifx0

    1. f(3)

    2. f(0)

    3. f(4)

  3. 55. g(x)={x+3ifx3(x+3)ifx<3

    1. g(0)

    2. g(6)

    3. g(3)

  4. 56. g(x)={x+5ifx5(x+5)ifx<5

    1. g(0)

    2. g(6)

    3. g(5)

  5. 57. h(x)={x29x3ifx36ifx=3

    1. h(5)

    2. h(0)

    3. h(3)

  6. 58. h(x)={x225x5ifx510ifx=5

    1. h(7)

    2. h(0)

    3. h(5)

In Exercises 5970, the domain of each piecewise function is (, ).

  1. Graph each function.

  2. Use your graph to determine the function’s range.

  1. 59. f(x)={xifx<0xifx0

  2. 60. f(x)={xifx<0xifx0

  3. 61. f(x)={2xifx02ifx>0

  4. 62. f(x)={12 xifx03ifx>0

  5. 63. f(x)={x+3ifx<2x3ifx2

  6. 64. f(x)={x+2ifx<3x2ifx3

  7. 65. f(x)={3ifx13ifx>1

  8. 66. f(x)={4ifx14ifx>1

  9. 67. f(x)={12 x2ifx<12x1ifx1

  10. 68. f(x)={ 12 x2ifx<12x+1ifx1

  11. 69. f(x)={0ifx<4xif4x<0x2ifx0

  12. 70. f(x)={0ifx<3 xif3x<0x21ifx0

In Exercises 7192, find and simplify the difference quotient

f(x+h)f(x)h, h0

for the given function.

  1. 71. f(x)=4x

  2. 72. f(x)=7x

  3. 73. f(x)=3x+7

  4. 74. f(x)=6x+1

  5. 75. f(x)=x2

  6. 76. f(x)=2x2

  7. 77. f(x)=x24x+3

  8. 78. f(x)=x25x+8

  9. 79. f(x)=2x2+x1

  10. 80. f(x)=3x2+x+5

  11. 81. f(x)=x2+2x+4

  12. 82. f(x)=x23x+1

  13. 83. f(x)=2x2+5x+7

  14. 84. f(x)=3x2+2x1

  15. 85. f(x)=2x2x+3

  16. 86. f(x)=3x2+x1

  17. 87. f(x)=6

  18. 88. f(x)=7

  19. 89. f(x)=1x

  20. 90. f(x)=12x

  21. 91. f(x)=x

  22. 92. f(x)=x1

Practice PLUS

In Exercises 9394, let f be defined by the following graph:

The image shows a stepwise function graph.
  1. 93. Find

    f(1.5)+f(0.9)[f(π)]2+f(3)÷f(1)f(π).
  2. 94. Find

    f(2.5)f(1.9)[f(π)]2+f(3)÷f(1)f(π).

A cable company offers the following high-speed Internet plans. Also given are the piecewise functions that model these plans. Use this information to solve Exercises 9596.

Plan A

  • $40 per month includes 400 GB of data.

  • Additional data costs $0.20 per GB.

    C(g)={40if0g40040+0.20(g400)ifg>400

Plan B

  • $60 per month includes 1000 GB of data.

  • Additional data costs $0.20 per GB.

    C(g)={60if0g100060+0.20(g1000)ifg>1000
  1. 95. Simplify the algebraic expression in the second line of the piecewise function for plan A. Then use point-plotting to graph the function.

  2. 96. Simplify the algebraic expression in the second line of the piecewise function for plan B. Then use point-plotting to graph the function.

In Exercises 9798, write a piecewise function that models each high-speed Internet plan. Then graph the function.

  1. 97. $50 per month includes 600 GB. Additional data costs $0.30 per GB.

  2. 98. $80 per month includes 2000 GB. Additional data costs $0.35 per GB.

Application Exercises

With aging, body fat increases and muscle mass declines. The line graphs show the percent body fat in adult women and men as they age from 25 to 75 years. Use the graphs to solve Exercises 99106.

A graph plots the percent of body fat in adults.

Source: Thompson et al., The Science of Nutrition, Benjamin Cummings, 2008

  1. 99. State the intervals on which the graph giving the percent body fat in women is increasing and decreasing.

  2. 100. State the intervals on which the graph giving the percent body fat in men is increasing and decreasing.

  3. 101. For what age does the percent body fat in women reach a maximum? What is the percent body fat for that age?

  4. 102. At what age does the percent body fat in men reach a maximum? What is the percent body fat for that age?

  5. 103. Use interval notation to give the domain and the range for the graph of the function for women.

  6. 104. Use interval notation to give the domain and the range for the graph of the function for men.

  7. 105. The function p(x)=0.002x2+0.15x+22.86 models percent body fat, p(x), where x is the number of years a person’s age exceeds 25. Use the graphs to determine whether this model describes percent body fat in women or in men.

  8. 106. The function p(x)=0.004x2+0.25x+33.64 models percent body fat, p(x), where x is the number of years a person’s age exceeds 25. Use the graphs to determine whether this model describes percent body fat in women or in men.

Here is the Federal Tax Rate Schedule X that specifies the tax owed by a single taxpayer for 2020.

If Your Taxable Income Is Over But Not Over The Tax You Owe Is Of the Amount Over
$          0 $    9,875                       10% $          0
$    9875 $  40,125 $   987.50+12% $    9875
$  40,125 $  85,525 $   4617.50+22% $  40,125
$  85,525 $163,300 $14,605.50+24% $  85,525
$163,300 $207,350 $33,271.50+32% $163,300
$207,350 $518,400 $47,367.50+35% $207,350
$518,400       − $   156,235+37% $518,400

The preceding tax table can be modeled by a piecewise function, where x represents the taxable income of a single taxpayer and T(x) is the tax owed:

T(x)={0.10xif0<x9875987.50+0.12(x9875)if9875<x40,1254617.50+0.22(x40,125)if40,125<x85,52514,605.50+0.24(x85,525)if85,525<x163,30033,271.50+0.32(x163,300)if163,300<x207,350wewdwdwdss?sdwdwdwddw¯if207,350<x518,400sssasassxxax?xssceddvdddd¯ifx>518,400

Use this information to solve Exercises 107108.

  1. 107. Find and interpret T(35,000).

  2. 108. Find and interpret T(50,000).

In Exercises 109110, refer to the preceding tax table.

  1. 109. Find the algebraic expression for the missing piece of T(x) that models tax owed for the domain (207,350,518,400].

  2. 110. Find the algebraic expression for the missing piece of T(x) that models tax owed for the domain (518,400, ).

The figure shows the cost of mailing a first-class letter, f(x), as a function of its weight, x, in ounces, in June 2020. Use the graph to solve Exercises 111114.

The image shows a function graph that plots cost, in dollars, versus weight, in ounces.

Source: Lynn E. Baring, Postmaster, Inverness, CA

  1. 111. Find f(3). What does this mean in terms of the variables in this situation?

  2. 112. Find f(3.75). What does this mean in terms of the variables in this situation?

  3. 113. What is the cost of mailing a letter that weighs 1.5 ounces?

  4. 114. What is the cost of mailing a letter that weighs 1.8 ounces?

  5. 115. Furry Finances. A pet insurance policy has a monthly rate that is a function of the age of the insured dog or cat. For pets whose age does not exceed 4, the monthly cost is $20. The cost then increases by $2 for each successive year of the pet’s age.

    Age Not Exceeding Monthly Cost
    4 $20
    5 $22
    6 $24

    The cost schedule continues in this manner for ages not exceeding 10. The cost for pets whose ages exceed 10 is $40. Use this information to create a graph that shows the monthly cost of the insurance, f(x), for a pet of age x, where the function’s domain is [0, 14].

Explaining the Concepts

  1. 116. What does it mean if a function f is increasing on an interval?

  2. 117. Suppose that a function f whose graph contains no breaks or gaps on (a, c) is increasing on (a, b), decreasing on (b, c), and defined at b. Describe what occurs at x=b. What does the function value f(b) represent?

  3. 118. Given an equation in x and y, how do you determine if its graph is symmetric with respect to the y-axis?

  4. 119. Given an equation in x and y, how do you determine if its graph is symmetric with respect to the x-axis?

  5. 120. Given an equation in x and y, how do you determine if its graph is symmetric with respect to the origin?

  6. 121. If you are given a function’s graph, how do you determine if the function is even, odd, or neither?

  7. 122. If you are given a function’s equation, how do you determine if the function is even, odd, or neither?

  8. 123. What is a piecewise function?

  9. 124. Explain how to find the difference quotient of a function f, f(x+h)f(x)h, if an equation for f is given.

Technology Exercises

  1. 125. The function

    f(x)=0.00002x3+0.008x20.3x+6.95

    models the number of annual physician visits, f(x), by a person of age x. Graph the function in a [0,100,5] by [0,40,2] viewing rectangle. What does the shape of the graph indicate about the relationship between one’s age and the number of annual physician visits? Use the  TABLE  or minimum function capability to find the coordinates of the minimum point on the graph of the function. What does this mean?

In Exercises 126131, use a graphing utility to graph each function. Use a [5, 5, 1] by [5, 5, 1] viewing rectangle. Then find the intervals on which the function is increasing, decreasing, or constant.

  1. 126. f(x)=x36x2+9x+1

  2. 127. g(x)=|4x2|

  3. 128. h(x)=|x2|+|x+2|

  4. 129. f(x)=x13(x4)

  5. 130. g(x)=x23

  6. 131. h(x)=2x25

  7. 132.

    1. Graph the functions f(x)=xn for n=2, 4, and 6 in a [2, 2, 1] by [1, 3, 1] viewing rectangle.

    2. Graph the functions f(x)=xn for n=1, 3, and 5 in a [2, 2, 1] by [2, 2, 1] viewing rectangle.

    3. If n is positive and even, where is the graph of f(x)=xn increasing and where is it decreasing?

    4. If n is positive and odd, what can you conclude about the graph of f(x)=xn in terms of increasing or decreasing behavior?

    5. Graph all six functions in a [1, 3, 1] by [1, 3, 1] viewing rectangle. What do you observe about the graphs in terms of how flat or how steep they are?

Critical Thinking Exercises

Make Sense? In Exercises 133136, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 133. My graph is decreasing on (, a) and increasing on (a, ), so f(a) must be a relative maximum.

  2. 134. This work by artist Scott Kim has the same kind of symmetry as an even function.

    The word Dyslexia is written such that the alphabets on either side of the letter, l, appear as a mirror image of each other. Double prime DYSLEXIA comma double prime 19 81.
  3. 135. I graphed

    f(x)={2ifx43ifx=4

    and one piece of my graph is a single point.

  4. 136. I noticed that the difference quotient is always zero if f(x)=c, where c is any constant.

  5. 137. Sketch the graph of f using the following properties. (More than one correct graph is possible.) f is a piecewise function that is decreasing on (, 2), f(2)=0, f is increasing on (2, ), and the range of f is [0, ).

  6. 138. Define a piecewise function on the intervals (, 2], (2, 5), and [5, ) that does not “jump” at 2 or 5 such that one piece is a constant function, another piece is an increasing function, and the third piece is a decreasing function.

  7. 139. Suppose that h(x)=f(x)g(x). The function f can be even, odd, or neither. The same is true for the function g.

    1. Under what conditions is h definitely an even function?

    2. Under what conditions is h definitely an odd function?

Retaining the Concepts

  1. 140. You invested $80,000 in two accounts paying 1.5% and 1.7% annual interest. If the total interest earned for the year was $1320, how much was invested at each rate? (Section P.8, Example 5)

  2. 141. Solve for A: C=A+Ar.

    (Section P.7, Example 5)

  3. 142. Solve by the quadratic formula: 5x26x8=0.

    (Section P.7, Example 10)

Preview Exercises

Exercises 143145 will help you prepare for the material covered in the next section.

  1. 143. If (x1, y1)=(3, 1) and (x2, y2)=(2, 4), find y2y1x2x1.

  2. 144. Find the ordered pairs (_______, 0) and (0, _______) satisfying 4x3y6=0.

  3. 145. Solve for y: 3x+2y4=0.

1.3: Exercise Set

1.3 Exercise Set

Practice Exercises

In Exercises 112, use the graph to determine

  1. intervals on which the function is increasing, if any.

  2. intervals on which the function is decreasing, if any.

  3. intervals on which the function is constant, if any.

  1. 1.

    A graph is an upward opening parabola with indefinite ends that falls from (negative 4, 5) through (negative 3, 0) to its vertex at (negative 1, negative 4), and then rises through (1, 0) to point (4, 4).
  2. 2.

    A graph is a downward opening parabola with indefinite ends that rises from (negative 4, negative 5) through (negative 3, 0) to its vertex at (negative 1, 4) and falls through (1, 0) to point (2, negative 5). All values are estimated.
  3. 3.

    A graph is a curve with a closed point and indefinite end at (0, 1) and passes through (4, 3). All values are estimated.
  4. 4.

    A graph is a curve with a closed point and indefinite end that starts at (negative 1, 0) and passes through (0, 1) and (3, 2). All values are estimated.
  5. 5.

    A graph is a line with closed points that falls from (negative 2, 6), passes through (0, 4), (4, 0), (0, 4) and ends at (6, negative 2). All values are estimated.
  6. 6.

    A graph is a line with closed points that rises from (negative 3, negative 5) (negative 0.5, 0), passes through (0, 1), and ends at (2, 5). All values are estimated.
  7. 7.

    A graph is a line that rises through (negative 4, negative 5) to (negative 1, 2) and then extends horizontally through (0, negative 2) and (4, negative 2). All values are estimated.
  8. 8.

    A graph is a line that passes along the negative x axis to the origin and then rises through (1, 2) and (2, 4). All values are estimated.
  9. 9.

    A graph is an m shaped curve with indefinite ends that start in the third quadrant and end in the fourth quadrant.
  10. 10.

    A graph is a sinusoidal curve with closed ends that starts at (negative 5, 0), passes through (negative 4, 4), (negative 3, 0), (negative 2, negative 4), (negative 1, 0), (0, 4), (1, 0), (2, negative 4), (3, 0), (4, 4), and ends at (5, 0).
  11. 11.

    A graph is a line curve with indefinite ends that passes through (negative 3, negative 2), (negative 2, negative 2), (0, negative 1), (2, 0), (4, 2), and (5, 2).
  12. 12.

    A graph is a curve with indefinite ends that passes through (negative 5, 2), (negative 4, 2), (0, 0), (2, negative 2), and (3, negative 2).

In Exercises 1316, the graph of a function f is given. Use the graph to find each of the following:

  1. The numbers, if any, at which f has a relative maximum. What are these relative maxima?

  2. The numbers, if any, at which f has a relative minimum. What are these relative minima?

  1. 13.

    A graph is a w shaped curve with indefinite ends that passes through (negative 4, 3), (negative 3, 0), (0, 4), (3, 0), and (4, 3).
  2. 14.

    A graph is a curve that passes through (negative 3, negative 1), (0, 2), and (3, negative 1).
  3. 15.

    A graphing calculator screen displays a N shaped curve, f of x = 2 x cubed + 3 x squared minus 12 x + 1, which passes through the third quadrant, second quadrant, fourth quadrant, and then first quadrant.

  4. 16.

    A graphing calculator screen displays a curve, f of x = 2 x cubed minus 15 x squared + 24 x + 19, which passes through the third, second, and first quadrant.

In Exercises 1732, determine whether the graph of each equation is symmetric with respect to the y-axis, the x-axis, the origin, more than one of these, or none of these.

  1. 17. y=x2+6

  2. 18. y=x22

  3. 19. x=y2+6

  4. 20. x=y22

  5. 21. y2=x2+6

  6. 22. y2=x22

  7. 23. y=2x+3

  8. 24. y=2x+5

  9. 25. x2y3=2

  10. 26. x3y2=5

  11. 27. x2+y2=100

  12. 28. x2+y2=49

  13. 29. x2y2+3xy=1

  14. 30. x2y2+5xy=2

  15. 31. y4=x3+6

  16. 32. y5=x4+2

In Exercises 3336, use possible symmetry to determine whether each graph is the graph of an even function, an odd function, or a function that is neither even nor odd.

  1. 33.

    A graph is a curve with indefinite ends that passes through (negative 2, 4 fifths), (0, 4), and (2, 4 fifths).
  2. 34.

    A graph plots two curves with indefinite ends. The first curve passes through (1, 4), (2, 2), and (4, 1). The second curve passes through (negative 4, negative 1), (negative 2, negative 2), and (negative 1, negative 4).
  3. 35.

    A graph is a curve that passes through (negative 1, 1), (0, 0), and (1, negative 1).
  4. 36.

    A graph is a curve that passes through (negative 1, 3), (0, 2), (1, 1), and (2, negative 5).

In Exercises 3748, determine whether each function is even, odd, or neither. Then determine whether the function’s graph is symmetric with respect to the y-axis, the origin, or neither.

  1. 37. f(x)=x3+x

  2. 38. f(x)=x3x

  3. 39. g(x)=x2+x

  4. 40. g(x)=x2x

  5. 41. h(x)=x2x4

  6. 42. h(x)=2x2+x4

  7. 43. f(x)=x2x4+1

  8. 44. f(x)=2x2+x4+1

  9. 45. f(x)=15 x63x2

  10. 46. f(x)=2x36x5

  11. 47. f(x)=x1x2

  12. 48. f(x)=x21x2

  13. 49. Use the graph of f to determine each of the following. Where applicable, use interval notation.

    The graph of y = f of x is a horizontal line with an indefinite end from (negative 1, 4) to (0, 4) and a curve with an indefinite end that passes through (0, 4), (1, 0), (4, negative 4), and (7, 0).
    1. the domain of f

    2. the range of f

    3. the x-intercepts

    4. the y-intercept

    5. intervals on which f is increasing

    6. intervals on which f is decreasing

    7. intervals on which f is constant

    8. the number at which f has a relative minimum

    9. the relative minimum of f

    10. f(3)

    11. the values of x for which f(x)=2

    12. Is f even, odd, or neither?

  14. 50. Use the graph of f to determine each of the following. Where applicable, use interval notation.

    The graph of y = f of x is a curve with indefinite ends that passes through (negative 4, 0), (negative 2, 4), (0, 1), (3, 2), and (4, 0).
    1. the domain of f

    2. the range of f

    3. the x-intercepts

    4. the y-intercept

    5. intervals on which f is increasing

    6. intervals on which f is decreasing

    7. values of x for which f(x)0

    8. the numbers at which f has a relative maximum

    9. the relative maxima of f

    10. f(2)

    11. the values of x for which f(x)=0

    12. Is f even, odd, or neither?

  15. 51. Use the graph of f to determine each of the following. Where applicable, use interval notation.

    The graph of y = f of x is a curve with indefinite ends that starts at (3, 0), passes through (1, 4), (0, 3), and (negative 3, 0).
    1. the domain of f

    2. the range of f

    3. the zeros of f

    4. f(0)

    5. intervals on which f is increasing

    6. intervals on which f is decreasing

    7. values of x for which f(x)0

    8. any relative maxima and the numbers at which they occur

    9. the value of x for which f(x)=4

    10. Is f(1) positive or negative?

  16. 52. Use the graph of f to determine each of the following. Where applicable, use interval notation.

    The graph of y = f of x is a line that rises through (negative 5, negative 2) to (negative 2, 1), a horizontal line from (negative 2, 1) to (2, 1), and a line that falls to a closed point (5, negative 2).
    1. the domain of f

    2. the range of f

    3. the zeros of f

    4. f(0)

    5. intervals on which f is increasing

    6. intervals on which f is decreasing

    7. intervals on which f is constant

    8. values of x for which f(x)>0

    9. values of x for which f(x)=2

    10. Is f(4) positive or negative?

    11. Is f even, odd, or neither?

    12. Is f(2) a relative maximum?

In Exercises 5358, evaluate each piecewise function at the given values of the independent variable.

  1. 53. f(x)={3x+5ifx<04x+7ifx0

    1. f(2)

    2. f(0)

    3. f(3)

  2. 54. f(x)={6x1ifx<07x+3ifx0

    1. f(3)

    2. f(0)

    3. f(4)

  3. 55. g(x)={x+3ifx3(x+3)ifx<3

    1. g(0)

    2. g(6)

    3. g(3)

  4. 56. g(x)={x+5ifx5(x+5)ifx<5

    1. g(0)

    2. g(6)

    3. g(5)

  5. 57. h(x)={x29x3ifx36ifx=3

    1. h(5)

    2. h(0)

    3. h(3)

  6. 58. h(x)={x225x5ifx510ifx=5

    1. h(7)

    2. h(0)

    3. h(5)

In Exercises 5970, the domain of each piecewise function is (, ).

  1. Graph each function.

  2. Use your graph to determine the function’s range.

  1. 59. f(x)={xifx<0xifx0

  2. 60. f(x)={xifx<0xifx0

  3. 61. f(x)={2xifx02ifx>0

  4. 62. f(x)={12 xifx03ifx>0

  5. 63. f(x)={x+3ifx<2x3ifx2

  6. 64. f(x)={x+2ifx<3x2ifx3

  7. 65. f(x)={3ifx13ifx>1

  8. 66. f(x)={4ifx14ifx>1

  9. 67. f(x)={12 x2ifx<12x1ifx1

  10. 68. f(x)={ 12 x2ifx<12x+1ifx1

  11. 69. f(x)={0ifx<4xif4x<0x2ifx0

  12. 70. f(x)={0ifx<3 xif3x<0x21ifx0

In Exercises 7192, find and simplify the difference quotient

f(x+h)f(x)h, h0

for the given function.

  1. 71. f(x)=4x

  2. 72. f(x)=7x

  3. 73. f(x)=3x+7

  4. 74. f(x)=6x+1

  5. 75. f(x)=x2

  6. 76. f(x)=2x2

  7. 77. f(x)=x24x+3

  8. 78. f(x)=x25x+8

  9. 79. f(x)=2x2+x1

  10. 80. f(x)=3x2+x+5

  11. 81. f(x)=x2+2x+4

  12. 82. f(x)=x23x+1

  13. 83. f(x)=2x2+5x+7

  14. 84. f(x)=3x2+2x1

  15. 85. f(x)=2x2x+3

  16. 86. f(x)=3x2+x1

  17. 87. f(x)=6

  18. 88. f(x)=7

  19. 89. f(x)=1x

  20. 90. f(x)=12x

  21. 91. f(x)=x

  22. 92. f(x)=x1

Practice PLUS

In Exercises 9394, let f be defined by the following graph:

The image shows a stepwise function graph.
  1. 93. Find

    f(1.5)+f(0.9)[f(π)]2+f(3)÷f(1)f(π).
  2. 94. Find

    f(2.5)f(1.9)[f(π)]2+f(3)÷f(1)f(π).

A cable company offers the following high-speed Internet plans. Also given are the piecewise functions that model these plans. Use this information to solve Exercises 9596.

Plan A

  • $40 per month includes 400 GB of data.

  • Additional data costs $0.20 per GB.

    C(g)={40if0g40040+0.20(g400)ifg>400

Plan B

  • $60 per month includes 1000 GB of data.

  • Additional data costs $0.20 per GB.

    C(g)={60if0g100060+0.20(g1000)ifg>1000
  1. 95. Simplify the algebraic expression in the second line of the piecewise function for plan A. Then use point-plotting to graph the function.

  2. 96. Simplify the algebraic expression in the second line of the piecewise function for plan B. Then use point-plotting to graph the function.

In Exercises 9798, write a piecewise function that models each high-speed Internet plan. Then graph the function.

  1. 97. $50 per month includes 600 GB. Additional data costs $0.30 per GB.

  2. 98. $80 per month includes 2000 GB. Additional data costs $0.35 per GB.

Application Exercises

With aging, body fat increases and muscle mass declines. The line graphs show the percent body fat in adult women and men as they age from 25 to 75 years. Use the graphs to solve Exercises 99106.

A graph plots the percent of body fat in adults.

Source: Thompson et al., The Science of Nutrition, Benjamin Cummings, 2008

  1. 99. State the intervals on which the graph giving the percent body fat in women is increasing and decreasing.

  2. 100. State the intervals on which the graph giving the percent body fat in men is increasing and decreasing.

  3. 101. For what age does the percent body fat in women reach a maximum? What is the percent body fat for that age?

  4. 102. At what age does the percent body fat in men reach a maximum? What is the percent body fat for that age?

  5. 103. Use interval notation to give the domain and the range for the graph of the function for women.

  6. 104. Use interval notation to give the domain and the range for the graph of the function for men.

  7. 105. The function p(x)=0.002x2+0.15x+22.86 models percent body fat, p(x), where x is the number of years a person’s age exceeds 25. Use the graphs to determine whether this model describes percent body fat in women or in men.

  8. 106. The function p(x)=0.004x2+0.25x+33.64 models percent body fat, p(x), where x is the number of years a person’s age exceeds 25. Use the graphs to determine whether this model describes percent body fat in women or in men.

Here is the Federal Tax Rate Schedule X that specifies the tax owed by a single taxpayer for 2020.

If Your Taxable Income Is Over But Not Over The Tax You Owe Is Of the Amount Over
$          0 $    9,875                       10% $          0
$    9875 $  40,125 $   987.50+12% $    9875
$  40,125 $  85,525 $   4617.50+22% $  40,125
$  85,525 $163,300 $14,605.50+24% $  85,525
$163,300 $207,350 $33,271.50+32% $163,300
$207,350 $518,400 $47,367.50+35% $207,350
$518,400       − $   156,235+37% $518,400

The preceding tax table can be modeled by a piecewise function, where x represents the taxable income of a single taxpayer and T(x) is the tax owed:

T(x)={0.10xif0<x9875987.50+0.12(x9875)if9875<x40,1254617.50+0.22(x40,125)if40,125<x85,52514,605.50+0.24(x85,525)if85,525<x163,30033,271.50+0.32(x163,300)if163,300<x207,350wewdwdwdss?sdwdwdwddw¯if207,350<x518,400sssasassxxax?xssceddvdddd¯ifx>518,400

Use this information to solve Exercises 107108.

  1. 107. Find and interpret T(35,000).

  2. 108. Find and interpret T(50,000).

In Exercises 109110, refer to the preceding tax table.

  1. 109. Find the algebraic expression for the missing piece of T(x) that models tax owed for the domain (207,350,518,400].

  2. 110. Find the algebraic expression for the missing piece of T(x) that models tax owed for the domain (518,400, ).

The figure shows the cost of mailing a first-class letter, f(x), as a function of its weight, x, in ounces, in June 2020. Use the graph to solve Exercises 111114.

The image shows a function graph that plots cost, in dollars, versus weight, in ounces.

Source: Lynn E. Baring, Postmaster, Inverness, CA

  1. 111. Find f(3). What does this mean in terms of the variables in this situation?

  2. 112. Find f(3.75). What does this mean in terms of the variables in this situation?

  3. 113. What is the cost of mailing a letter that weighs 1.5 ounces?

  4. 114. What is the cost of mailing a letter that weighs 1.8 ounces?

  5. 115. Furry Finances. A pet insurance policy has a monthly rate that is a function of the age of the insured dog or cat. For pets whose age does not exceed 4, the monthly cost is $20. The cost then increases by $2 for each successive year of the pet’s age.

    Age Not Exceeding Monthly Cost
    4 $20
    5 $22
    6 $24

    The cost schedule continues in this manner for ages not exceeding 10. The cost for pets whose ages exceed 10 is $40. Use this information to create a graph that shows the monthly cost of the insurance, f(x), for a pet of age x, where the function’s domain is [0, 14].

Explaining the Concepts

  1. 116. What does it mean if a function f is increasing on an interval?

  2. 117. Suppose that a function f whose graph contains no breaks or gaps on (a, c) is increasing on (a, b), decreasing on (b, c), and defined at b. Describe what occurs at x=b. What does the function value f(b) represent?

  3. 118. Given an equation in x and y, how do you determine if its graph is symmetric with respect to the y-axis?

  4. 119. Given an equation in x and y, how do you determine if its graph is symmetric with respect to the x-axis?

  5. 120. Given an equation in x and y, how do you determine if its graph is symmetric with respect to the origin?

  6. 121. If you are given a function’s graph, how do you determine if the function is even, odd, or neither?

  7. 122. If you are given a function’s equation, how do you determine if the function is even, odd, or neither?

  8. 123. What is a piecewise function?

  9. 124. Explain how to find the difference quotient of a function f, f(x+h)f(x)h, if an equation for f is given.

Technology Exercises

  1. 125. The function

    f(x)=0.00002x3+0.008x20.3x+6.95

    models the number of annual physician visits, f(x), by a person of age x. Graph the function in a [0,100,5] by [0,40,2] viewing rectangle. What does the shape of the graph indicate about the relationship between one’s age and the number of annual physician visits? Use the  TABLE  or minimum function capability to find the coordinates of the minimum point on the graph of the function. What does this mean?

In Exercises 126131, use a graphing utility to graph each function. Use a [5, 5, 1] by [5, 5, 1] viewing rectangle. Then find the intervals on which the function is increasing, decreasing, or constant.

  1. 126. f(x)=x36x2+9x+1

  2. 127. g(x)=|4x2|

  3. 128. h(x)=|x2|+|x+2|

  4. 129. f(x)=x13(x4)

  5. 130. g(x)=x23

  6. 131. h(x)=2x25

  7. 132.

    1. Graph the functions f(x)=xn for n=2, 4, and 6 in a [2, 2, 1] by [1, 3, 1] viewing rectangle.

    2. Graph the functions f(x)=xn for n=1, 3, and 5 in a [2, 2, 1] by [2, 2, 1] viewing rectangle.

    3. If n is positive and even, where is the graph of f(x)=xn increasing and where is it decreasing?

    4. If n is positive and odd, what can you conclude about the graph of f(x)=xn in terms of increasing or decreasing behavior?

    5. Graph all six functions in a [1, 3, 1] by [1, 3, 1] viewing rectangle. What do you observe about the graphs in terms of how flat or how steep they are?

Critical Thinking Exercises

Make Sense? In Exercises 133136, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 133. My graph is decreasing on (, a) and increasing on (a, ), so f(a) must be a relative maximum.

  2. 134. This work by artist Scott Kim has the same kind of symmetry as an even function.

    The word Dyslexia is written such that the alphabets on either side of the letter, l, appear as a mirror image of each other. Double prime DYSLEXIA comma double prime 19 81.
  3. 135. I graphed

    f(x)={2ifx43ifx=4

    and one piece of my graph is a single point.

  4. 136. I noticed that the difference quotient is always zero if f(x)=c, where c is any constant.

  5. 137. Sketch the graph of f using the following properties. (More than one correct graph is possible.) f is a piecewise function that is decreasing on (, 2), f(2)=0, f is increasing on (2, ), and the range of f is [0, ).

  6. 138. Define a piecewise function on the intervals (, 2], (2, 5), and [5, ) that does not “jump” at 2 or 5 such that one piece is a constant function, another piece is an increasing function, and the third piece is a decreasing function.

  7. 139. Suppose that h(x)=f(x)g(x). The function f can be even, odd, or neither. The same is true for the function g.

    1. Under what conditions is h definitely an even function?

    2. Under what conditions is h definitely an odd function?

Retaining the Concepts

  1. 140. You invested $80,000 in two accounts paying 1.5% and 1.7% annual interest. If the total interest earned for the year was $1320, how much was invested at each rate? (Section P.8, Example 5)

  2. 141. Solve for A: C=A+Ar.

    (Section P.7, Example 5)

  3. 142. Solve by the quadratic formula: 5x26x8=0.

    (Section P.7, Example 10)

Preview Exercises

Exercises 143145 will help you prepare for the material covered in the next section.

  1. 143. If (x1, y1)=(3, 1) and (x2, y2)=(2, 4), find y2y1x2x1.

  2. 144. Find the ordered pairs (_______, 0) and (0, _______) satisfying 4x3y6=0.

  3. 145. Solve for y: 3x+2y4=0.

Section 1.4: Linear Functions and Slope

Section 1.4 Linear Functions and Slope

Learning Objectives

What You’ll Learn

  1. 1 Calculate a line’s slope.

  2. 2 Write the point-slope form of the equation of a line.

  3. 3 Write and graph the slope-intercept form of the equation of a line.

  4. 4 Graph horizontal or vertical lines.

  5. 5 Recognize and use the general form of a line’s equation.

  6. 6 Use intercepts to graph the general form of a line’s equation.

  7. 7 Model data with linear functions and make predictions.

Is there a relationship between literacy and child mortality? As the percentage of adult females who are literate increases, does the mortality of children under five decrease? Figure 1.40 indicates that this is, indeed, the case. Each point in the figure represents one country.

Figure 1.40

A scatterplot tiled, Literacy and Child Mortality, plots under five mortality, per thousand, versus the percentage of adult females who are literate.

Source: United Nations

Figure 1.40 Full Alternative Text

Data presented in a visual form as a set of points is called a scatter plot. Also shown in Figure 1.40 is a line that passes through or near the points. A line that best fits the data points in a scatter plot is called a regression line. By writing the equation of this line, we can obtain a model for the data and make predictions about child mortality based on the percentage of literate adult females in a country.

Data often fall on or near a line. In this section, we will use functions to model such data and make predictions. We begin with a discussion of a line’s steepness.

Objective 1: Calculate a line’s slope

The Slope of a Line

  1. Objective 1 Calculate a line’s slope.

Watch Video

Mathematicians have developed a useful measure of the steepness of a line, called the slope of the line. Slope compares the vertical change (the rise) to the horizontal change (the run) when moving from one fixed point to another along the line. To calculate the slope of a line, we use a ratio that compares the change in y (the rise) to the corresponding change in x (the run).

Definition of Slope

The slope of the line through the distinct points (x1, y1) and (x2, y2) is

Start fraction change in y over change in x end fraction = start fraction rise labeled, vertical change, overrun labeled, horizontal change, end fraction = start fraction y sub 2 minus y sub 1 over x sub 2 minus x sub 1 end fraction.
A graph plots a line that rises from the third quadrant through the second quadrant, and the points, (x sub 1, y sub 1) and (x sub 2, y sub 2), in the first quadrant.

where x2x10.

It is common notation to let the letter m represent the slope of a line. The letter m is used because it is the first letter of the French verb monter, meaning “to rise” or “to ascend.”

Example 1 Using the Definition of Slope

Find the slope of the line passing through each pair of points:

  1. (3, 1) and (2, 4)

  2. (3, 4) and (2, 2).

Solution

  1. Let (x1, y1)=(3, 1) and (x2, y2)=(2, 4). We obtain the slope as follows:

    m=Change inyChange inx=y2y1x2x1=4(1)2(3)=4+12+3=51=5.

    The situation is illustrated in Figure 1.41. The slope of the line is 5. For every vertical change, or rise, of 5 units, there is a corresponding horizontal change, or run, of 1 unit. The slope is positive and the line rises from left to right.

    Figure 1.41 Visualizing a slope of 5

    A graph plots a line that rises through the points (negative 4, negative 6), (negative 3, negative 1), and (negative 2, 4).
  2. To find the slope of the line passing through (3, 4) and (2, 2), we can let (x1, y1)=(3, 4) and (x2, y2)=(2, 2). The slope of the line is computed as follows:

    m=Change inyChange inx=y2y1x2x1=242(3)=65= 65.

    The situation is illustrated in Figure 1.42. The slope of the line is 65. For every vertical change of 6 units (6 units down), there is a corresponding horizontal change of 5 units. The slope is negative and the line falls from left to right.

    Figure 1.42 Visualizing a slope of  65

    A graph plots a line with indefinite ends that falls through the points, (negative 3, 4), (0, 0), and (2, negative 2).

Check Point 1

  • Find the slope of the line passing through each pair of points:

    1. (3, 4) and (4, 2)

    2. (4, 2) and (1, 5).

Example 1 illustrates that a line with a positive slope is increasing and a line with a negative slope is decreasing. By contrast, a horizontal line is a constant function and has a slope of zero. A vertical line has no horizontal change, so x2x1=0 in the formula for slope. Because we cannot divide by zero, the slope of a vertical line is undefined. This discussion is summarized in Table 1.3.

Table 1.3 Possibilities for a Line’s Slope

The image shows four graphs that depict the types of slope.
Table 1.3 Full Alternative Text
Objective 1: Calculate a line’s slope

The Slope of a Line

  1. Objective 1 Calculate a line’s slope.

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Mathematicians have developed a useful measure of the steepness of a line, called the slope of the line. Slope compares the vertical change (the rise) to the horizontal change (the run) when moving from one fixed point to another along the line. To calculate the slope of a line, we use a ratio that compares the change in y (the rise) to the corresponding change in x (the run).

Definition of Slope

The slope of the line through the distinct points (x1, y1) and (x2, y2) is

Start fraction change in y over change in x end fraction = start fraction rise labeled, vertical change, overrun labeled, horizontal change, end fraction = start fraction y sub 2 minus y sub 1 over x sub 2 minus x sub 1 end fraction.
A graph plots a line that rises from the third quadrant through the second quadrant, and the points, (x sub 1, y sub 1) and (x sub 2, y sub 2), in the first quadrant.

where x2x10.

It is common notation to let the letter m represent the slope of a line. The letter m is used because it is the first letter of the French verb monter, meaning “to rise” or “to ascend.”

Example 1 Using the Definition of Slope

Find the slope of the line passing through each pair of points:

  1. (3, 1) and (2, 4)

  2. (3, 4) and (2, 2).

Solution

  1. Let (x1, y1)=(3, 1) and (x2, y2)=(2, 4). We obtain the slope as follows:

    m=Change inyChange inx=y2y1x2x1=4(1)2(3)=4+12+3=51=5.

    The situation is illustrated in Figure 1.41. The slope of the line is 5. For every vertical change, or rise, of 5 units, there is a corresponding horizontal change, or run, of 1 unit. The slope is positive and the line rises from left to right.

    Figure 1.41 Visualizing a slope of 5

    A graph plots a line that rises through the points (negative 4, negative 6), (negative 3, negative 1), and (negative 2, 4).
  2. To find the slope of the line passing through (3, 4) and (2, 2), we can let (x1, y1)=(3, 4) and (x2, y2)=(2, 2). The slope of the line is computed as follows:

    m=Change inyChange inx=y2y1x2x1=242(3)=65= 65.

    The situation is illustrated in Figure 1.42. The slope of the line is 65. For every vertical change of 6 units (6 units down), there is a corresponding horizontal change of 5 units. The slope is negative and the line falls from left to right.

    Figure 1.42 Visualizing a slope of  65

    A graph plots a line with indefinite ends that falls through the points, (negative 3, 4), (0, 0), and (2, negative 2).

Check Point 1

  • Find the slope of the line passing through each pair of points:

    1. (3, 4) and (4, 2)

    2. (4, 2) and (1, 5).

Example 1 illustrates that a line with a positive slope is increasing and a line with a negative slope is decreasing. By contrast, a horizontal line is a constant function and has a slope of zero. A vertical line has no horizontal change, so x2x1=0 in the formula for slope. Because we cannot divide by zero, the slope of a vertical line is undefined. This discussion is summarized in Table 1.3.

Table 1.3 Possibilities for a Line’s Slope

The image shows four graphs that depict the types of slope.
Table 1.3 Full Alternative Text
Objective 2: Write the point-slope form of the equation of a line

The Point-Slope Form of the Equation of a Line

  1. Objective 2 Write the point-slope form of the equation of a line.

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We can use the slope of a line to obtain various forms of the line’s equation. For example, consider a nonvertical line that has slope m and that contains the point (x1, y1).

The line in Figure 1.43 has slope m and contains the point (x1, y1). Let (x, y) represent any other point on the line.

Figure 1.43 A line passing through (x1, y1) with slope m

A graph is a line that rises through (x sub 1, y sub 1) and (x, y). The slope of a line between the two points is m with run = x minus x sub 1 and rise = y minus y sub 1. The point (x sub 1, y sub 1) is fixed, and the point (x, y) is arbitrary.

Regardless of where the point (x, y) is located, the steepness of the line in Figure 1.43 remains the same. Thus, the ratio for the slope stays a constant m. This means that for all points (x, y) along the line

The image shows the process of calculating the slope of the line.
1.4-188 Full Alternative Text

We can clear the fraction by multiplying both sides by xx1, the least common denominator.

m=yy1xx1This is the slope of the line in Figure 1.43.m(xx1)=yy1xx1(xx1)Multiply both sides by xx1.m(xx1)=yy1Simplify: yy1xx1(xx1)=yy1.

Now, if we reverse the two sides, we obtain the point-slope form of the equation of a line.

Point-Slope Form of the Equation of a Line

The point-slope form of the equation of a nonvertical line with slope m that passes through the point (x1, y1) is

yy1=m(xx1).

For example, the point-slope form of the equation of the line passing through (1, 5) with slope 2 (m=2) is

y5=2(x1).

We will soon be expressing the equation of a nonvertical line in function notation. To do so, we need to solve the point-slope form of a line’s equation for y. Example 2 illustrates how to isolate y on one side of the equal sign.

Example 2 Writing an Equation for a Line in Point-Slope Form

Write an equation in point-slope form for the line with slope 4 that passes through the point (1, 3). Then solve the equation for y.

Solution

We use the point-slope form of the equation of a line with m=4, x1=1, and y1=3.

yy1=m(xx1)This is the point-slope form of the equation. y3=4[x(1)]Substitute the given values: m=4 and (x1,y1)=(1,3).y3=4(x+1)We now have an equation in point-slope form for the given line.

Now we solve this equation for y.

The image shows a mathematical expression to find the value of y.

Check Point 2

  • Write an equation in point-slope form for the line with slope 6 that passes through the point (2, 5). Then solve the equation for y.

Example 3 Writing an Equation for a Line in Point-Slope Form

Write an equation in point-slope form for the line passing through the points (4, 3) and (2, 6). (See Figure 1.44.) Then solve the equation for y.

Figure 1.44 Write an equation in point-slope form for this line.

A graph is a line with indefinite ends that falls through (negative 2, 6), (0, 3), (2, 0), and (4, negative 3).

Solution

To use the point-slope form, we need to find the slope. The slope is the change in the y-coordinates divided by the corresponding change in the x-coordinates.

m=6(3)24=96= 32This is the definition of slope using (4,3) and (2,6).

We can take either point on the line to be (x1, y1). Let’s use (x1, y1)=(4, 3). Now, we are ready to write the point-slope form of the equation.

yy1=m(xx1)This is the point-slope form of the equation.y(3)=32(x4)Substitute:(x1,y1)=(4,3) and m=32.y+3=32(x4)Simplify.

We now have an equation in point-slope form for the line shown in Figure 1.44. Now, we solve this equation for y.

The image shows a mathematical expression to find the value of y.

Check Point 3

  • Write an equation in point-slope form for the line passing through the points (2, 1) and (1, 6). Then solve the equation for y.

Objective 2: Write the point-slope form of the equation of a line

The Point-Slope Form of the Equation of a Line

  1. Objective 2 Write the point-slope form of the equation of a line.

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We can use the slope of a line to obtain various forms of the line’s equation. For example, consider a nonvertical line that has slope m and that contains the point (x1, y1).

The line in Figure 1.43 has slope m and contains the point (x1, y1). Let (x, y) represent any other point on the line.

Figure 1.43 A line passing through (x1, y1) with slope m

A graph is a line that rises through (x sub 1, y sub 1) and (x, y). The slope of a line between the two points is m with run = x minus x sub 1 and rise = y minus y sub 1. The point (x sub 1, y sub 1) is fixed, and the point (x, y) is arbitrary.

Regardless of where the point (x, y) is located, the steepness of the line in Figure 1.43 remains the same. Thus, the ratio for the slope stays a constant m. This means that for all points (x, y) along the line

The image shows the process of calculating the slope of the line.
1.4-188 Full Alternative Text

We can clear the fraction by multiplying both sides by xx1, the least common denominator.

m=yy1xx1This is the slope of the line in Figure 1.43.m(xx1)=yy1xx1(xx1)Multiply both sides by xx1.m(xx1)=yy1Simplify: yy1xx1(xx1)=yy1.

Now, if we reverse the two sides, we obtain the point-slope form of the equation of a line.

Point-Slope Form of the Equation of a Line

The point-slope form of the equation of a nonvertical line with slope m that passes through the point (x1, y1) is

yy1=m(xx1).

For example, the point-slope form of the equation of the line passing through (1, 5) with slope 2 (m=2) is

y5=2(x1).

We will soon be expressing the equation of a nonvertical line in function notation. To do so, we need to solve the point-slope form of a line’s equation for y. Example 2 illustrates how to isolate y on one side of the equal sign.

Example 2 Writing an Equation for a Line in Point-Slope Form

Write an equation in point-slope form for the line with slope 4 that passes through the point (1, 3). Then solve the equation for y.

Solution

We use the point-slope form of the equation of a line with m=4, x1=1, and y1=3.

yy1=m(xx1)This is the point-slope form of the equation. y3=4[x(1)]Substitute the given values: m=4 and (x1,y1)=(1,3).y3=4(x+1)We now have an equation in point-slope form for the given line.

Now we solve this equation for y.

The image shows a mathematical expression to find the value of y.

Check Point 2

  • Write an equation in point-slope form for the line with slope 6 that passes through the point (2, 5). Then solve the equation for y.

Example 3 Writing an Equation for a Line in Point-Slope Form

Write an equation in point-slope form for the line passing through the points (4, 3) and (2, 6). (See Figure 1.44.) Then solve the equation for y.

Figure 1.44 Write an equation in point-slope form for this line.

A graph is a line with indefinite ends that falls through (negative 2, 6), (0, 3), (2, 0), and (4, negative 3).

Solution

To use the point-slope form, we need to find the slope. The slope is the change in the y-coordinates divided by the corresponding change in the x-coordinates.

m=6(3)24=96= 32This is the definition of slope using (4,3) and (2,6).

We can take either point on the line to be (x1, y1). Let’s use (x1, y1)=(4, 3). Now, we are ready to write the point-slope form of the equation.

yy1=m(xx1)This is the point-slope form of the equation.y(3)=32(x4)Substitute:(x1,y1)=(4,3) and m=32.y+3=32(x4)Simplify.

We now have an equation in point-slope form for the line shown in Figure 1.44. Now, we solve this equation for y.

The image shows a mathematical expression to find the value of y.

Check Point 3

  • Write an equation in point-slope form for the line passing through the points (2, 1) and (1, 6). Then solve the equation for y.

Objective 3: Write and graph the slope-intercept form of the equation of a line

The Slope-Intercept Form of the Equation of a Line

  1. Objective 3 Write and graph the slope-intercept form of the equation of a line.

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Let’s write the point-slope form of the equation of a nonvertical line with slope m and y-intercept b. The line is shown in Figure 1.45. Because the y-intercept is b, the line passes through (0, b). We use the point-slope form with x1=0 and y1=b.

y minus y sub 1 = m left parenthesis x minus x sub 1 right parenthesis. Let y sub 1 = b and let x sub 1 = 0.

Figure 1.45 A line with slope m and y-intercept b

A graph is a line that falls from a point on the y axis (0, b) to a point near the x axis in the first quadrant (x, y).
Figure 1.45 Full Alternative Text

We obtain

yb=m(x0).

Simplifying on the right side gives us

yb=mx.

Finally, we solve for y by adding b to both sides.

y=mx+b

Thus, if a line’s equation is written as y=mx+b with y isolated on one side, the coefficient of x is the line’s slope and the constant term is the y-intercept. This form of a line’s equation is called the slope-intercept form of the line.

Slope-Intercept Form of the Equation of a Line

The slope-intercept form of the equation of a nonvertical line with slope m and y-intercept b is

y=mx+b.

The slope-intercept form of a line’s equation, y=mx+b, can be expressed in function notation by replacing y with f(x):

f(x)=mx+b.

We have seen that functions in this form are called linear functions. Thus, in the equation of a linear function, the coefficient of x is the line’s slope and the constant term is the y-intercept. Here are two examples:

y = 2 x minus 4. 2 is labeled, the slope is 2. 4 is labeled, the y intercept is negative 4. f of x = 1 half x + 2. 1 half is labeled, the slope is 1 half. 2 is labeled, the y intercept is 2.

If a linear function’s equation is in slope-intercept form, we can use the y-intercept and the slope to obtain its graph.

Graphing y=mx+b Using the Slope and y-Intercept

  1. Plot the point containing the y-intercept on the y-axis. This is the point (0, b).

  2. Obtain a second point using the slope, m. Write m as a fraction, and use rise over run, starting at the point containing the y-intercept, to plot this point.

  3. Use a straightedge to draw a line through the two points. Draw arrowheads at the ends of the line to show that the line continues indefinitely in both directions.

Example 4 Graphing Using the Slope and y-Intercept

Graph the linear function: f(x)= 32 x+2.

Solution

The equation of the line is in the form f(x)=mx+b. We can find the slope, m, by identifying the coefficient of x. We can find the y-intercept, b, by identifying the constant term.

f of x = negative start fraction 3 over 2 end fraction x + 2. Negative start fraction 3 over 2 end fraction is labeled, the slope is negative start fraction 3 over 2 end fraction. 2 is labeled, the y intercept is 2.

Now that we have identified the slope, 32, and the y-intercept, 2, we use the three-step procedure to graph the equation.

  1. Step 1 PLOT THE POINT CONTAINING THE Y-INTERCEPT ON THE Y-AXIS. The y-intercept is 2. We plot (0,2), shown in Figure 1.46.

    Figure 1.46 The graph of f(x)= 32 x+2

    A graph plots a line that falls through (negative 2, 5) (0, 2), and (2, negative 1). The y intercept is 2. The slope of the line between (0, 2) and (2, negative 1) has rise = negative 3 and run = 2.
  2. Step 2 OBTAIN A SECOND POINT USING THE SLOPE, M. WRITE M AS A FRACTION, AND USE RISE OVER RUN, STARTING AT THE POINT CONTAINING THE Y-INTERCEPT, TO PLOT THIS POINT. The slope, 32, is already written as a fraction.

    m= 32=32=RiseRun

    We plot the second point on the line by starting at (0,2), the first point. Based on the slope, we move 3 units down (the rise) and 2 units to the right (the run). This puts us at a second point on the line, (2, 1), shown in Figure 1.46.

  3. Step 3 USE A STRAIGHTEDGE TO DRAW A LINE THROUGH THE TWO POINTS. The graph of the linear function f(x)= 32 x+2 is shown as a blue line in Figure 1.46.

Check Point 4

  • Graph the linear function: f(x)=35 x+1.

Objective 3: Write and graph the slope-intercept form of the equation of a line

The Slope-Intercept Form of the Equation of a Line

  1. Objective 3 Write and graph the slope-intercept form of the equation of a line.

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Let’s write the point-slope form of the equation of a nonvertical line with slope m and y-intercept b. The line is shown in Figure 1.45. Because the y-intercept is b, the line passes through (0, b). We use the point-slope form with x1=0 and y1=b.

y minus y sub 1 = m left parenthesis x minus x sub 1 right parenthesis. Let y sub 1 = b and let x sub 1 = 0.

Figure 1.45 A line with slope m and y-intercept b

A graph is a line that falls from a point on the y axis (0, b) to a point near the x axis in the first quadrant (x, y).
Figure 1.45 Full Alternative Text

We obtain

yb=m(x0).

Simplifying on the right side gives us

yb=mx.

Finally, we solve for y by adding b to both sides.

y=mx+b

Thus, if a line’s equation is written as y=mx+b with y isolated on one side, the coefficient of x is the line’s slope and the constant term is the y-intercept. This form of a line’s equation is called the slope-intercept form of the line.

Slope-Intercept Form of the Equation of a Line

The slope-intercept form of the equation of a nonvertical line with slope m and y-intercept b is

y=mx+b.

The slope-intercept form of a line’s equation, y=mx+b, can be expressed in function notation by replacing y with f(x):

f(x)=mx+b.

We have seen that functions in this form are called linear functions. Thus, in the equation of a linear function, the coefficient of x is the line’s slope and the constant term is the y-intercept. Here are two examples:

y = 2 x minus 4. 2 is labeled, the slope is 2. 4 is labeled, the y intercept is negative 4. f of x = 1 half x + 2. 1 half is labeled, the slope is 1 half. 2 is labeled, the y intercept is 2.

If a linear function’s equation is in slope-intercept form, we can use the y-intercept and the slope to obtain its graph.

Graphing y=mx+b Using the Slope and y-Intercept

  1. Plot the point containing the y-intercept on the y-axis. This is the point (0, b).

  2. Obtain a second point using the slope, m. Write m as a fraction, and use rise over run, starting at the point containing the y-intercept, to plot this point.

  3. Use a straightedge to draw a line through the two points. Draw arrowheads at the ends of the line to show that the line continues indefinitely in both directions.

Example 4 Graphing Using the Slope and y-Intercept

Graph the linear function: f(x)= 32 x+2.

Solution

The equation of the line is in the form f(x)=mx+b. We can find the slope, m, by identifying the coefficient of x. We can find the y-intercept, b, by identifying the constant term.

f of x = negative start fraction 3 over 2 end fraction x + 2. Negative start fraction 3 over 2 end fraction is labeled, the slope is negative start fraction 3 over 2 end fraction. 2 is labeled, the y intercept is 2.

Now that we have identified the slope, 32, and the y-intercept, 2, we use the three-step procedure to graph the equation.

  1. Step 1 PLOT THE POINT CONTAINING THE Y-INTERCEPT ON THE Y-AXIS. The y-intercept is 2. We plot (0,2), shown in Figure 1.46.

    Figure 1.46 The graph of f(x)= 32 x+2

    A graph plots a line that falls through (negative 2, 5) (0, 2), and (2, negative 1). The y intercept is 2. The slope of the line between (0, 2) and (2, negative 1) has rise = negative 3 and run = 2.
  2. Step 2 OBTAIN A SECOND POINT USING THE SLOPE, M. WRITE M AS A FRACTION, AND USE RISE OVER RUN, STARTING AT THE POINT CONTAINING THE Y-INTERCEPT, TO PLOT THIS POINT. The slope, 32, is already written as a fraction.

    m= 32=32=RiseRun

    We plot the second point on the line by starting at (0,2), the first point. Based on the slope, we move 3 units down (the rise) and 2 units to the right (the run). This puts us at a second point on the line, (2, 1), shown in Figure 1.46.

  3. Step 3 USE A STRAIGHTEDGE TO DRAW A LINE THROUGH THE TWO POINTS. The graph of the linear function f(x)= 32 x+2 is shown as a blue line in Figure 1.46.

Check Point 4

  • Graph the linear function: f(x)=35 x+1.

Objective 4: Graph horizontal or vertical lines

Equations of Horizontal and Vertical Lines

  1. Objective 4 Graph horizontal or vertical lines.

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If a line is horizontal, its slope is zero: m=0. Thus, the equation y=mx+b becomes y=b, where b is the y-intercept. All horizontal lines have equations of the form y=b.

Example 5 Graphing a Horizontal Line

Graph y=4 in the rectangular coordinate system.

A table lists the values of x and y.

Solution

All ordered pairs that are solutions of y=4 have a value of y that is always 4. Any value can be used for x. In the table on the right, we have selected three of the possible values for x: 2, 0, and 3. The table shows that three ordered pairs that are solutions of y=4 are (2, 4), (0, 4), and (3, 4). Drawing a line that passes through the three points gives the horizontal line shown in Figure 1.47.

Figure 1.47 The graph of y=4 or f(x)=4

A graph plots a horizontal line with indefinite ends parallel to the x axis that passes through (negative 2, negative 4), (0, negative 4), and (3, negative 4). The y intercept is negative 4.

Check Point 5

  • Graph y=3 in the rectangular coordinate system.

Equation of a Horizontal Line

A horizontal line is given by an equation of the form

y=b,

where b is the y-intercept of the line. The slope of a horizontal line is zero.

A graph plots a horizontal line with indefinite ends parallel to the x axis that passes through the second quadrant intersecting the positive y axis at (0, b) towards the first quadrant. The point of intersection is labeled, y intercept, b.

Because any vertical line can intersect the graph of a horizontal line y=b only once, a horizontal line is the graph of a function. Thus, we can express the equation y=b as f(x)=b. This linear function is often called a constant function.

Next, let’s see what we can discover about the graph of an equation of the form x=a by looking at an example.

Example 6 Graphing a Vertical Line

Graph the linear equation: x=2.

A table lists the values of x and y.

Solution

All ordered pairs that are solutions of x=2 have a value of x that is always 2. Any value can be used for y. In the table on the right, we have selected three of the possible values for y: 2, 0, and 3. The table shows that three ordered pairs that are solutions of x=2 are (2, 2), (2, 0), and (2, 3). Drawing a line that passes through the three points gives the vertical line shown in Figure 1.48.

Figure 1.48 The graph of x=2

A graph plots a vertical line with indefinite ends parallel to the y axis that passes through (2, 3), (2, 0), and (2, negative 2). The x intercept is 2.

Does a vertical line represent the graph of a linear function? No. Look at the graph of x=2 in Figure 1.48. A vertical line drawn through (2, 0) intersects the graph infinitely many times. This shows that infinitely many outputs are associated with the input 2. No vertical line represents a linear function.

Equation of a Vertical Line

A vertical line is given by an equation of the form

x=a,

where a is the x-intercept of the line. The slope of a vertical line is undefined.

A graph plots a vertical line with indefinite ends parallel to the y axis that passes through the first quadrant intersecting through the positive x axis at (a, 0) towards the fourth quadrant.

Check Point 6

  • Graph the linear equation: x=3.

Objective 5: Recognize and use the general form of a line’s equation

The General Form of the Equation of a Line

  1. Objective 5 Recognize and use the general form of a line’s equation.

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The vertical line whose equation is x=5 cannot be written in slope-intercept form, y=mx+b, because its slope is undefined. However, every line has an equation that can be expressed in the form Ax+By+C=0. For example, x=5 can be expressed as 1x+0y5=0, or x5=0. The equation Ax+By+C=0 is called the general form of the equation of a line.

General Form of the Equation of a Line

Every line has an equation that can be written in the general form

Ax+By+C=0,

where A, B, and C are real numbers, and A and B are not both zero.

If the equation of a nonvertical line is given in general form, it is possible to find the slope, m, and the y-intercept, b, for the line. We solve the equation for y, transforming it into the slope-intercept form y=mx+b. In this form, the coefficient of x is the slope of the line and the constant term is its y-intercept.

Example 7 Finding the Slope and the y-Intercept

Find the slope and the y-intercept of the line whose equation is 3x+2y4=0.

Solution

The equation is given in general form. We begin by rewriting it in the form y=mx+b. We need to solve for y.

The image shows a mathematical expression to find the value of y.

The coefficient of x,  32, is the slope and the constant term, 2, is the y-intercept. This is the form of the equation that we graphed in Figure 1.46.

Check Point 7

  • Find the slope and the y-intercept of the line whose equation is 3x+6y12=0. Then use the y-intercept and the slope to graph the equation.

Objective 6: Use intercepts to graph the general form of a line’s equation

Using Intercepts to Graph Ax + By + C = 0

  1. Objective 6 Use intercepts to graph the general form of a line’s equation.

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Example 7 and Check Point 7 illustrate that one way to graph the general form of a line’s equation is to convert to slope-intercept form, y=mx+b. Then use the slope and the y-intercept to obtain the graph.

A second method for graphing Ax+By+C=0 uses intercepts. This method does not require rewriting the general form in a different form.

Using Intercepts to Graph Ax+By+C=0

  1. Find the x-intercept. Let y=0 and solve for x. Plot the point containing the x-intercept on the x-axis.

  2. Find the y-intercept. Let x=0 and solve for y. Plot the point containing the y-intercept on the y-axis.

  3. Use a straightedge to draw a line through the two points containing the intercepts. Draw arrowheads at the ends of the line to show that the line continues indefinitely in both directions.

As long as none of A, B, and C is zero, the graph of Ax+By+C=0 will have distinct x- and y-intercepts, and this three-step method can be used to graph the equation.

Example 8 Using Intercepts to Graph a Linear Equation

Graph using intercepts: 4x3y6=0.

Solution

  1. Step 1 FIND THE x-INTERCEPT. LET y=0 AND SOLVE FOR x.

    4x306=0Replace y with 0 in 4x3y6=0.4x6=0Simplify.4x=6Add 6 to both sides.x=64=32Divide both sides by 4.

    The x-intercept is 32, so the line passes through (32, 0) or (1.5, 0), as shown in Figure 1.49.

    Figure 1.49 The graph of 4x3y6=0

    A graph is a line that rises through the third, fourth and first quadrant, intersecting at points (0, negative 2) and (start fraction 3 over 2 end fraction, 0). x intercept is start fraction 3 over 2 end fraction, and the y intercept is negative 2.
  2. Step 2 FIND THE y-INTERCEPT. LET x=0 AND SOLVE FOR y.

    403y6=0Replace x with 0 in 4x3y6=0.3y6=0Simplify.3y=6Add 6 to both sides.y=2Divide both sides by 3.

    The y-intercept is 2, so the line passes through (0, 2), as shown in Figure 1.49.

  3. Step 3 GRAPH THE EQUATION BY DRAWING A LINE THROUGH THE TWO POINTS CONTAINING THE INTERCEPTS. The graph of 4x3y6=0 is shown in Figure 1.49.

Check Point 8

  • Graph using intercepts: 3x2y6=0.

We’ve covered a lot of territory. Let’s take a moment to summarize the various forms for equations of lines.

Equations of Lines

1. Point-slope form yy1=m(xx1)
2. Slope-intercept form y=mx+b or f(x)=mx+b
3. Horizontal line y=b or f(x)=b
4. Vertical line x=a
5. General form Ax+By+C=0
Objective 7: Model data with linear functions and make predictions

Applications

  1. Objective 7 Model data with linear functions and make predictions.

Watch Video

Linear functions are useful for modeling data that fall on or near a line.

Example 9 Modeling Global Warming

The amount of carbon dioxide in the atmosphere, measured in parts per million, has been increasing as a result of the burning of oil and coal. The buildup of gases and particles traps heat and raises the planet’s temperature. The bar graph in Figure 1.50(a) on the next page gives the average atmospheric concentration of carbon dioxide and the average global temperature for six selected years. The data are displayed as a set of six points in a rectangular coordinate system in Figure 1.50(b).

Figure 1.50(a)(b)

Two graphs titled average global temperature.

Source: National Oceanic and Atmospheric Administration

  1. Shown on the scatter plot in Figure 1.50(b) is a line that passes through or near the six points. Write the slope-intercept form of this equation using function notation.

  2. The preindustrial concentration of atmospheric carbon dioxide was 280 parts per million. The United Nations’ Intergovernmental Panel on Climate Change predicts global temperatures will rise between 2°F and 5°F if carbon dioxide concentration doubles from the preindustrial level. Compared to the average global temperature of 57.99°F for 2009, how well does the function from part (a) model this prediction?

Solution

  1. The line in Figure 1.50(b) passes through (326,57.06) and (385,57.99). We start by finding its slope.

    m=Change inyChange inx=57.9957.06385326=0.93590.02

    The slope indicates that for each increase of one part per million in carbon dioxide concentration, the average global temperature is increasing by approximately 0.02°F.

    Now we write the line’s equation in slope-intercept form.

    yy1=m(xx1)Begin with the point-slope form.y57.06=0.02(x326)Either ordered pair can be (x1,y1).Let (x1,y1)=(326,57.06).From above,m0.02.y57.06=0.02x6.52Apply the distributive property:0.02(326)=6.52.y=0.02x+50.54Add 57.06 to both sides and solve for y.

    A linear function that models average global temperature, f(x), for an atmospheric carbon dioxide concentration of x parts per million is

    f(x)=0.02x+50.54.
  2. If carbon dioxide concentration doubles from its preindustrial level of 280 parts per million, which many experts deem very likely, the concentration will reach 280×2, or 560 parts per million. We use the linear function to predict average global temperature at this concentration.

    f(x)=0.02x+50.54Use the function from part (a).f(560)=0.02(560)+50.54Substitute 560 for x.=11.2+50.54=61.74

    Our model projects an average global temperature of 61.74°F for a carbon dioxide concentration of 560 parts per million. Compared to the average global temperature of 57.99°F for 2009 shown in Figure 1.50(a) on the previous page, this is an increase of

    61.74°F57.99°F=3.75°F.

    This is consistent with a rise between 2°F and 5°F as predicted by the Intergovernmental Panel on Climate Change.

Check Point 9

  • Use the data points (317,57.04) and (354,57.64), shown, but not labeled, in Figure 1.50(b) to obtain a linear function that models average global temperature, f(x), for an atmospheric carbon dioxide concentration of x parts per million. Round m to three decimal places and b to one decimal place. Then use the function to project average global temperature at a concentration of 600 parts per million.

Objective 7: Model data with linear functions and make predictions

Applications

  1. Objective 7 Model data with linear functions and make predictions.

Watch Video

Linear functions are useful for modeling data that fall on or near a line.

Example 9 Modeling Global Warming

The amount of carbon dioxide in the atmosphere, measured in parts per million, has been increasing as a result of the burning of oil and coal. The buildup of gases and particles traps heat and raises the planet’s temperature. The bar graph in Figure 1.50(a) on the next page gives the average atmospheric concentration of carbon dioxide and the average global temperature for six selected years. The data are displayed as a set of six points in a rectangular coordinate system in Figure 1.50(b).

Figure 1.50(a)(b)

Two graphs titled average global temperature.

Source: National Oceanic and Atmospheric Administration

  1. Shown on the scatter plot in Figure 1.50(b) is a line that passes through or near the six points. Write the slope-intercept form of this equation using function notation.

  2. The preindustrial concentration of atmospheric carbon dioxide was 280 parts per million. The United Nations’ Intergovernmental Panel on Climate Change predicts global temperatures will rise between 2°F and 5°F if carbon dioxide concentration doubles from the preindustrial level. Compared to the average global temperature of 57.99°F for 2009, how well does the function from part (a) model this prediction?

Solution

  1. The line in Figure 1.50(b) passes through (326,57.06) and (385,57.99). We start by finding its slope.

    m=Change inyChange inx=57.9957.06385326=0.93590.02

    The slope indicates that for each increase of one part per million in carbon dioxide concentration, the average global temperature is increasing by approximately 0.02°F.

    Now we write the line’s equation in slope-intercept form.

    yy1=m(xx1)Begin with the point-slope form.y57.06=0.02(x326)Either ordered pair can be (x1,y1).Let (x1,y1)=(326,57.06).From above,m0.02.y57.06=0.02x6.52Apply the distributive property:0.02(326)=6.52.y=0.02x+50.54Add 57.06 to both sides and solve for y.

    A linear function that models average global temperature, f(x), for an atmospheric carbon dioxide concentration of x parts per million is

    f(x)=0.02x+50.54.
  2. If carbon dioxide concentration doubles from its preindustrial level of 280 parts per million, which many experts deem very likely, the concentration will reach 280×2, or 560 parts per million. We use the linear function to predict average global temperature at this concentration.

    f(x)=0.02x+50.54Use the function from part (a).f(560)=0.02(560)+50.54Substitute 560 for x.=11.2+50.54=61.74

    Our model projects an average global temperature of 61.74°F for a carbon dioxide concentration of 560 parts per million. Compared to the average global temperature of 57.99°F for 2009 shown in Figure 1.50(a) on the previous page, this is an increase of

    61.74°F57.99°F=3.75°F.

    This is consistent with a rise between 2°F and 5°F as predicted by the Intergovernmental Panel on Climate Change.

Check Point 9

  • Use the data points (317,57.04) and (354,57.64), shown, but not labeled, in Figure 1.50(b) to obtain a linear function that models average global temperature, f(x), for an atmospheric carbon dioxide concentration of x parts per million. Round m to three decimal places and b to one decimal place. Then use the function to project average global temperature at a concentration of 600 parts per million.

1.4: Exercise Set

1.4 Exercise Set

Practice Exercises

In Exercises 110, find the slope of the line passing through each pair of points or state that the slope is undefined. Then indicate whether the line through the points rises, falls, is horizontal, or is vertical.

  1. 1. (4, 7) and (8, 10)

  2. 2. (2, 1) and (3, 4)

  3. 3. (2, 1) and (2, 2)

  4. 4. (1, 3) and (2, 4)

  5. 5. (4, 2) and (3, 2)

  6. 6. (4, 1) and (3, 1)

  7. 7. (2, 4) and (1, 1)

  8. 8. (6, 4) and (4, 2)

  9. 9. (5, 3) and (5, 2)

  10. 10. (3, 4) and (3, 5)

In Exercises 1138, use the given conditions to write an equation for each line in point-slope form and slope-intercept form.

  1. 11. Slope=2, passing through (3, 5)

  2. 12. Slope=4, passing through (1, 3)

  3. 13. Slope=6, passing through (2, 5)

  4. 14. Slope=8, passing through (4, 1)

  5. 15. Slope=3, passing through (2, 3)

  6. 16. Slope=5, passing through (4, 2)

  7. 17. Slope=4, passing through (4, 0)

  8. 18. Slope=2, passing through (0, 3)

  9. 19. Slope=1, passing through ( 12, 2)

  10. 20. Slope=1, passing through (4, 14)

  11. 21. Slope=12, passing through the origin

  12. 22. Slope=13, passing through the origin

  13. 23. Slope= 23, passing through (6, 2)

  14. 24. Slope= 35, passing through (10, 4)

  15. 25. Passing through (1, 2) and (5, 10)

  16. 26. Passing through (3, 5) and (8, 15)

  17. 27. Passing through (3, 0) and (0, 3)

  18. 28. Passing through (2, 0) and (0, 2)

  19. 29. Passing through (3, 1) and (2, 4)

  20. 30. Passing through (2, 4) and (1, 1)

  21. 31. Passing through (3, 2) and (3, 6)

  22. 32. Passing through (3, 6) and (3, 2)

  23. 33. Passing through (3, 1) and (4, 1)

  24. 34. Passing through (2, 5) and (6, 5)

  25. 35. Passing through (2, 4) with x-intercept=2

  26. 36. Passing through (1, 3) with x-intercept=1

  27. 37. x-intercept= 12 and y-intercept=4

  28. 38. x-intercept=4 and y-intercept=2

In Exercises 3948, give the slope and y-intercept of each line whose equation is given. Then graph the linear function.

  1. 39. y=2x+1

  2. 40. y=3x+2

  3. 41. f(x)=2x+1

  4. 42. f(x)=3x+2

  5. 43. f(x)=34 x2

  6. 44. f(x)=34 x3

  7. 45. y= 35 x+7

  8. 46. y= 25 x+6

  9. 47. g(x)= 12 x

  10. 48. g(x)= 13 x

In Exercises 4958, graph each equation in a rectangular coordinate system.

  1. 49. y=2

  2. 50. y=4

  3. 51. x=3

  4. 52. x=5

  5. 53. y=0

  6. 54. x=0

  7. 55. f(x)=1

  8. 56. f(x)=3

  9. 57. 3x18=0

  10. 58. 3x+12=0

In Exercises 5966,

  1. Rewrite the given equation in slope-intercept form.

  2. Give the slope and y-intercept.

  3. Use the slope and y-intercept to graph the linear function.

  1. 59. 3x+y5=0

  2. 60. 4x+y6=0

  3. 61. 2x+3y18=0

  4. 62. 4x+6y+12=0

  5. 63. 8x4y12=0

  6. 64. 6x5y20=0

  7. 65. 3y9=0

  8. 66. 4y+28=0

In Exercises 6772, use intercepts to graph each equation.

  1. 67. 6x2y12=0

  2. 68. 6x9y18=0

  3. 69. 2x+3y+6=0

  4. 70. 3x+5y+15=0

  5. 71. 8x2y+12=0

  6. 72. 6x3y+15=0

Practice PLUS

In Exercises 7376, find the slope of the line passing through each pair of points or state that the slope is undefined. Assume that all variables represent positive real numbers. Then indicate whether the line through the points rises, falls, is horizontal, or is vertical.

  1. 73. (0, a) and (b, 0)

  2. 74. (a, 0) and (0, b)

  3. 75. (a, b) and (a, b+c)

  4. 76. (ab, c) and (a, a+c)

In Exercises 7778, give the slope and y-intercept of each line whose equation is given. Assume that B0.

  1. 77. Ax+By=C

  2. 78. Ax=ByC

In Exercises 7980, find the value of y if the line through the two given points is to have the indicated slope.

  1. 79. (3, y) and (1, 4), m=3

  2. 80. (2, y) and (4, 4), m=13

In Exercises 8182, graph each linear function.

  1. 81. 3x4f(x)6=0

  2. 82. 6x5f(x)20=0

  3. 83. If one point on a line is (3, 1) and the line’s slope is 2, find the y-intercept.

  4. 84. If one point on a line is (2, 6) and the line’s slope is  32, find the y-intercept.

Use the figure to make the lists in Exercises 8586.

The image plots a graph of four lines.
  1. 85. List the slopes m1, m2, m3, and m4 in order of decreasing size.

  2. 86. List the y-intercepts b1, b2, b3, and b4 in order of decreasing size.

Application Exercises

Americans’ trust in government and the media has generally been on a downward trend since pollsters first asked questions on these topics in the second half of the twentieth century. Trust in government hit an all-time low of 14% in 2014, while trust in the media bottomed out at 32% in 2016. The bar graph shows the percentage of Americans trusting in the government and the media for five selected years. The data are displayed as two sets of five points each, one scatter plot for the percentage of Americans trusting in the government and one for the percentage of Americans trusting in the media. Also shown for each scatter plot is a line that passes through or near the five points. Use these lines to solve Exercises 8788.

The image shows two graphs titled, Americans trust in Government and media.

Sources: Gallup, Pew Research

  1. 87. In this exercise, you will use the red line for trust in government shown on the scatter plot to develop a model for the percentage of Americans trusting in government.

    1. Use the two points whose coordinates are shown by the voice balloons to find the point-slope form of the equation of the line that models the percentage of Americans trusting in government, y, x years after 2003. Round the slope to two decimal places.

    2. Write the equation from part (a) in slope-intercept form. Use function notation.

    3. Use the linear function to predict the percentage of Americans trusting in government in 2025. Round to the nearest percent.

  2. 88. In this exercise, you will use the blue line for trust in media shown on the scatter plot to develop a model for the percentage of Americans trusting in media.

    1. Use the two points whose coordinates are shown by the voice balloons to find the point-slope form of the equation of the line that models the percentage of Americans trusting in media, y, x years after 2003. Round the slope to two decimal places.

    2. Write the equation from part (a) in slope-intercept form. Use function notation.

    3. Use the linear function to predict the percentage of Americans trusting in media in 2025. Round to the nearest percent.

The bar graph gives the life expectancy for American men and women born in six selected years. In Exercises 8990, you will use the data to obtain models for life expectancy and make predictions about how long American men and women will live in the future.

The image shows a graph bar graph titled, life expectancy in the United States, by year of birth.

Source: National Center for Health Statistics

  1. 89. Use the data for males shown in the bar graph at the bottom of the previous column to solve this exercise.

    1. Let x represent the number of birth years after 1960 and let y represent male life expectancy. Create a scatter plot that displays the data as a set of six points in a rectangular coordinate system.

    2. Draw a line through the two points that show male life expectancies for 1980 and 2000. Use the coordinates of these points to write a linear function that models life expectancy, E(x), for American men born x years after 1960.

    3. Use the function from part (b) to project the life expectancy of American men born in 2020.

  2. 90. Use the data for females shown in the bar graph at the bottom of the previous column to solve this exercise.

    1. Let x represent the number of birth years after 1960 and let y represent female life expectancy. Create a scatter plot that displays the data as a set of six points in a rectangular coordinate system.

    2. Draw a line through the two points that show female life expectancies for 1970 and 2000. Use the coordinates of these points to write a linear function that models life expectancy, E(x), for American women born x years after 1960. Round the slope to two decimal places.

    3. Use the function from part (b) to project the life expectancy of American women born in 2020.

  3. 91. Shown, again, is the scatter plot that indicates a relationship between the percentage of adult females in a country who are literate and the mortality of children under five. Also shown is a line that passes through or near the points. Find a linear function that models the data by finding the slope-intercept form of the line’s equation. Use the function to make a prediction about child mortality based on the percentage of adult females in a country who are literate.

    The image shows a scatterplot titled literacy and child mortality.

    Source: United Nations

  4. 92. Just as money doesn’t buy happiness for individuals, the two don’t necessarily go together for countries either. However, the scatter plot does show a relationship between a country’s annual per capita income and the percentage of people in that country who call themselves “happy.”

    The image shows a graph titled, per capita income and national happiness.

    Source: Richard Layard, Happiness: Lessons from a New Science, Penguin, 2005

    Draw a line that fits the data so that the spread of the data points around the line is as small as possible. Use the coordinates of two points along your line to write the slope-intercept form of its equation. Express the equation in function notation and use the linear function to make a prediction about national happiness based on per capita income.

Explaining the Concepts

  1. 93. What is the slope of a line and how is it found?

  2. 94. Describe how to write the equation of a line if the coordinates of two points along the line are known.

  3. 95. Explain how to derive the slope-intercept form of a line’s equation, y=mx+b, from the point-slope form

    yy1=m(xx1).
  4. 96. Explain how to graph the equation x=2. Can this equation be expressed in slope-intercept form? Explain.

  5. 97. Explain how to use the general form of a line’s equation to find the line’s slope and y-intercept.

  6. 98. Explain how to use intercepts to graph the general form of a line’s equation.

  7. 99. Take another look at the scatter plot in Exercise 91. Although there is a relationship between literacy and child mortality, we cannot conclude that increased literacy causes child mortality to decrease. Offer two or more possible explanations for the data in the scatter plot.

Technology Exercises

Use a graphing utility to graph each equation in Exercises 100103. Then use the  TRACE  feature to trace along the line and find the coordinates of two points. Use these points to compute the line’s slope. Check your result by using the coefficient of x in the line’s equation.

  1. 100. y=2x+4

  2. 101. y=3x+6

  3. 102. y= 12 x5

  4. 103. y=34 x2

  5. 104. Is there a relationship between wine consumption and deaths from heart disease? The table gives data from 19 developed countries.

    A table lists data for the relation between wine consumption and deaths from heart disease.
    A table lists data for the relation between wine consumption and deaths from heart disease.

    Source: New York Times

    1. Use the statistical menu of your graphing utility to enter the 19 ordered pairs of data items shown in the table.

    2. Use the scatter plot capability to draw a scatter plot of the data.

    3. Select the linear regression option. Use your utility to obtain values for a and b for the equation of the regression line, y=ax+b. You may also be given a correlation coefficient, r. Values of r close to 1 indicate that the points can be described by a linear relationship and the regression line has a positive slope. Values of r close to 1 indicate that the points can be described by a linear relationship and the regression line has a negative slope. Values of r close to 0 indicate no linear relationship between the variables. In this case, a linear model does not accurately describe the data.

    4. Use the appropriate sequence (consult your manual) to graph the regression equation on top of the points in the scatter plot.

Critical Thinking Exercises

Make Sense? In Exercises 105108, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 105. The graph of my linear function at first increased, reached a maximum point, and then decreased.

  2. 106. A linear function that models tuition and fees at public four-year colleges from 2000 through 2020 has negative slope.

  3. 107. Because the variable m does not appear in Ax+By+C=0, equations in this form make it impossible to determine the line’s slope.

  4. 108. The federal minimum wage was $7.25 per hour from 2009 through 2020, so f(x)=7.25 models the minimum wage, f(x), in dollars, for the domain {2009, 2010, 2011, , 2020}.

In Exercises 109112, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 109. The equation y=mx+b shows that no line can have a y-intercept that is numerically equal to its slope.

  2. 110. Every line in the rectangular coordinate system has an equation that can be expressed in slope-intercept form.

  3. 111. The graph of the linear function 5x+6y30=0 is a line passing through the point (6, 0) with slope 56.

  4. 112. The graph of x=7 in the rectangular coordinate system is the single point (7, 0).

In Exercises 113114, find the coefficients that must be placed in each shaded area so that the function’s graph will be a line satisfying the specified conditions.

  1. 113.  x+ y12=0;x-intercept=2;y-intercept=4

  2. 114.  x+ y12=0;y-intercept=6;slope=12

  3. 115. Prove that the equation of a line passing through (a, 0) and (0, b)(a0, b0) can be written in the form xa+yb=1. Why is this called the intercept form of a line?

  4. 116. Excited about the success of celebrity stamps, post office officials were rumored to have put forth a plan to institute two new types of thermometers. On these new scales, °U represents degrees Usher and °R represents degrees Rihanna. If it is known that 40°U=25°R, 280°U=125°R, and degrees Usher is linearly related to degrees Rihanna, write an equation expressing U in terms of R.

Group Exercise

  1. 117. In Exercises 8788, we used the data in a bar graph to develop linear functions that modeled trust in government and trust in media. For this group exercise, you might find it helpful to pattern your work after Exercises 87 and 88. Group members should begin by consulting an almanac, newspaper, magazine, or the Internet to find data that appear to lie approximately on or near a line. Working by hand or using a graphing utility, group members should construct scatter plots for the data that were assembled. If working by hand, draw a line that approximately fits the data in each scatter plot and then write its equation as a function in slope-intercept form. If using a graphing utility, obtain the equation of each regression line. Then use each linear function’s equation to make predictions about what might occur in the future. Are there circumstances that might affect the accuracy of the prediction? List some of these circumstances.

Retaining the Concepts

  1. 118. According to the U.S. Office of Management and Budget, the cost of maintaining existing public transportation infrastructure in 2020 was $103.4 billion and is projected to increase by $0.9 billion each year. By which year is the cost of maintaining existing public transportation infrastructure expected to reach $116 billion?

    (Section P.8, Example 2)

In Exercises 119120, solve and graph the solution set on a number line.

  1. 119. x+34  x23+1

    (Section P.9, Example 5)

  2. 120. 3|2x+6|9<15

    (Section P.9, Example 8)

Preview Exercises

Exercises 121123 will help you prepare for the material covered in the next section.

  1. 121. Write the slope-intercept form of the equation of the line passing through (3, 1) whose slope is the same as the line whose equation is y=2x+1.

  2. 122. Write an equation in general form of the line passing through (3, 5) whose slope is the negative reciprocal (the reciprocal with the opposite sign) of 14.

  3. 123. If f(x)=x2, find

    f(x2)f(x1)x2x1,

    where x1=1 and x2=4.

1.4: Exercise Set

1.4 Exercise Set

Practice Exercises

In Exercises 110, find the slope of the line passing through each pair of points or state that the slope is undefined. Then indicate whether the line through the points rises, falls, is horizontal, or is vertical.

  1. 1. (4, 7) and (8, 10)

  2. 2. (2, 1) and (3, 4)

  3. 3. (2, 1) and (2, 2)

  4. 4. (1, 3) and (2, 4)

  5. 5. (4, 2) and (3, 2)

  6. 6. (4, 1) and (3, 1)

  7. 7. (2, 4) and (1, 1)

  8. 8. (6, 4) and (4, 2)

  9. 9. (5, 3) and (5, 2)

  10. 10. (3, 4) and (3, 5)

In Exercises 1138, use the given conditions to write an equation for each line in point-slope form and slope-intercept form.

  1. 11. Slope=2, passing through (3, 5)

  2. 12. Slope=4, passing through (1, 3)

  3. 13. Slope=6, passing through (2, 5)

  4. 14. Slope=8, passing through (4, 1)

  5. 15. Slope=3, passing through (2, 3)

  6. 16. Slope=5, passing through (4, 2)

  7. 17. Slope=4, passing through (4, 0)

  8. 18. Slope=2, passing through (0, 3)

  9. 19. Slope=1, passing through ( 12, 2)

  10. 20. Slope=1, passing through (4, 14)

  11. 21. Slope=12, passing through the origin

  12. 22. Slope=13, passing through the origin

  13. 23. Slope= 23, passing through (6, 2)

  14. 24. Slope= 35, passing through (10, 4)

  15. 25. Passing through (1, 2) and (5, 10)

  16. 26. Passing through (3, 5) and (8, 15)

  17. 27. Passing through (3, 0) and (0, 3)

  18. 28. Passing through (2, 0) and (0, 2)

  19. 29. Passing through (3, 1) and (2, 4)

  20. 30. Passing through (2, 4) and (1, 1)

  21. 31. Passing through (3, 2) and (3, 6)

  22. 32. Passing through (3, 6) and (3, 2)

  23. 33. Passing through (3, 1) and (4, 1)

  24. 34. Passing through (2, 5) and (6, 5)

  25. 35. Passing through (2, 4) with x-intercept=2

  26. 36. Passing through (1, 3) with x-intercept=1

  27. 37. x-intercept= 12 and y-intercept=4

  28. 38. x-intercept=4 and y-intercept=2

In Exercises 3948, give the slope and y-intercept of each line whose equation is given. Then graph the linear function.

  1. 39. y=2x+1

  2. 40. y=3x+2

  3. 41. f(x)=2x+1

  4. 42. f(x)=3x+2

  5. 43. f(x)=34 x2

  6. 44. f(x)=34 x3

  7. 45. y= 35 x+7

  8. 46. y= 25 x+6

  9. 47. g(x)= 12 x

  10. 48. g(x)= 13 x

In Exercises 4958, graph each equation in a rectangular coordinate system.

  1. 49. y=2

  2. 50. y=4

  3. 51. x=3

  4. 52. x=5

  5. 53. y=0

  6. 54. x=0

  7. 55. f(x)=1

  8. 56. f(x)=3

  9. 57. 3x18=0

  10. 58. 3x+12=0

In Exercises 5966,

  1. Rewrite the given equation in slope-intercept form.

  2. Give the slope and y-intercept.

  3. Use the slope and y-intercept to graph the linear function.

  1. 59. 3x+y5=0

  2. 60. 4x+y6=0

  3. 61. 2x+3y18=0

  4. 62. 4x+6y+12=0

  5. 63. 8x4y12=0

  6. 64. 6x5y20=0

  7. 65. 3y9=0

  8. 66. 4y+28=0

In Exercises 6772, use intercepts to graph each equation.

  1. 67. 6x2y12=0

  2. 68. 6x9y18=0

  3. 69. 2x+3y+6=0

  4. 70. 3x+5y+15=0

  5. 71. 8x2y+12=0

  6. 72. 6x3y+15=0

Practice PLUS

In Exercises 7376, find the slope of the line passing through each pair of points or state that the slope is undefined. Assume that all variables represent positive real numbers. Then indicate whether the line through the points rises, falls, is horizontal, or is vertical.

  1. 73. (0, a) and (b, 0)

  2. 74. (a, 0) and (0, b)

  3. 75. (a, b) and (a, b+c)

  4. 76. (ab, c) and (a, a+c)

In Exercises 7778, give the slope and y-intercept of each line whose equation is given. Assume that B0.

  1. 77. Ax+By=C

  2. 78. Ax=ByC

In Exercises 7980, find the value of y if the line through the two given points is to have the indicated slope.

  1. 79. (3, y) and (1, 4), m=3

  2. 80. (2, y) and (4, 4), m=13

In Exercises 8182, graph each linear function.

  1. 81. 3x4f(x)6=0

  2. 82. 6x5f(x)20=0

  3. 83. If one point on a line is (3, 1) and the line’s slope is 2, find the y-intercept.

  4. 84. If one point on a line is (2, 6) and the line’s slope is  32, find the y-intercept.

Use the figure to make the lists in Exercises 8586.

The image plots a graph of four lines.
  1. 85. List the slopes m1, m2, m3, and m4 in order of decreasing size.

  2. 86. List the y-intercepts b1, b2, b3, and b4 in order of decreasing size.

Application Exercises

Americans’ trust in government and the media has generally been on a downward trend since pollsters first asked questions on these topics in the second half of the twentieth century. Trust in government hit an all-time low of 14% in 2014, while trust in the media bottomed out at 32% in 2016. The bar graph shows the percentage of Americans trusting in the government and the media for five selected years. The data are displayed as two sets of five points each, one scatter plot for the percentage of Americans trusting in the government and one for the percentage of Americans trusting in the media. Also shown for each scatter plot is a line that passes through or near the five points. Use these lines to solve Exercises 8788.

The image shows two graphs titled, Americans trust in Government and media.

Sources: Gallup, Pew Research

  1. 87. In this exercise, you will use the red line for trust in government shown on the scatter plot to develop a model for the percentage of Americans trusting in government.

    1. Use the two points whose coordinates are shown by the voice balloons to find the point-slope form of the equation of the line that models the percentage of Americans trusting in government, y, x years after 2003. Round the slope to two decimal places.

    2. Write the equation from part (a) in slope-intercept form. Use function notation.

    3. Use the linear function to predict the percentage of Americans trusting in government in 2025. Round to the nearest percent.

  2. 88. In this exercise, you will use the blue line for trust in media shown on the scatter plot to develop a model for the percentage of Americans trusting in media.

    1. Use the two points whose coordinates are shown by the voice balloons to find the point-slope form of the equation of the line that models the percentage of Americans trusting in media, y, x years after 2003. Round the slope to two decimal places.

    2. Write the equation from part (a) in slope-intercept form. Use function notation.

    3. Use the linear function to predict the percentage of Americans trusting in media in 2025. Round to the nearest percent.

The bar graph gives the life expectancy for American men and women born in six selected years. In Exercises 8990, you will use the data to obtain models for life expectancy and make predictions about how long American men and women will live in the future.

The image shows a graph bar graph titled, life expectancy in the United States, by year of birth.

Source: National Center for Health Statistics

  1. 89. Use the data for males shown in the bar graph at the bottom of the previous column to solve this exercise.

    1. Let x represent the number of birth years after 1960 and let y represent male life expectancy. Create a scatter plot that displays the data as a set of six points in a rectangular coordinate system.

    2. Draw a line through the two points that show male life expectancies for 1980 and 2000. Use the coordinates of these points to write a linear function that models life expectancy, E(x), for American men born x years after 1960.

    3. Use the function from part (b) to project the life expectancy of American men born in 2020.

  2. 90. Use the data for females shown in the bar graph at the bottom of the previous column to solve this exercise.

    1. Let x represent the number of birth years after 1960 and let y represent female life expectancy. Create a scatter plot that displays the data as a set of six points in a rectangular coordinate system.

    2. Draw a line through the two points that show female life expectancies for 1970 and 2000. Use the coordinates of these points to write a linear function that models life expectancy, E(x), for American women born x years after 1960. Round the slope to two decimal places.

    3. Use the function from part (b) to project the life expectancy of American women born in 2020.

  3. 91. Shown, again, is the scatter plot that indicates a relationship between the percentage of adult females in a country who are literate and the mortality of children under five. Also shown is a line that passes through or near the points. Find a linear function that models the data by finding the slope-intercept form of the line’s equation. Use the function to make a prediction about child mortality based on the percentage of adult females in a country who are literate.

    The image shows a scatterplot titled literacy and child mortality.

    Source: United Nations

  4. 92. Just as money doesn’t buy happiness for individuals, the two don’t necessarily go together for countries either. However, the scatter plot does show a relationship between a country’s annual per capita income and the percentage of people in that country who call themselves “happy.”

    The image shows a graph titled, per capita income and national happiness.

    Source: Richard Layard, Happiness: Lessons from a New Science, Penguin, 2005

    Draw a line that fits the data so that the spread of the data points around the line is as small as possible. Use the coordinates of two points along your line to write the slope-intercept form of its equation. Express the equation in function notation and use the linear function to make a prediction about national happiness based on per capita income.

Explaining the Concepts

  1. 93. What is the slope of a line and how is it found?

  2. 94. Describe how to write the equation of a line if the coordinates of two points along the line are known.

  3. 95. Explain how to derive the slope-intercept form of a line’s equation, y=mx+b, from the point-slope form

    yy1=m(xx1).
  4. 96. Explain how to graph the equation x=2. Can this equation be expressed in slope-intercept form? Explain.

  5. 97. Explain how to use the general form of a line’s equation to find the line’s slope and y-intercept.

  6. 98. Explain how to use intercepts to graph the general form of a line’s equation.

  7. 99. Take another look at the scatter plot in Exercise 91. Although there is a relationship between literacy and child mortality, we cannot conclude that increased literacy causes child mortality to decrease. Offer two or more possible explanations for the data in the scatter plot.

Technology Exercises

Use a graphing utility to graph each equation in Exercises 100103. Then use the  TRACE  feature to trace along the line and find the coordinates of two points. Use these points to compute the line’s slope. Check your result by using the coefficient of x in the line’s equation.

  1. 100. y=2x+4

  2. 101. y=3x+6

  3. 102. y= 12 x5

  4. 103. y=34 x2

  5. 104. Is there a relationship between wine consumption and deaths from heart disease? The table gives data from 19 developed countries.

    A table lists data for the relation between wine consumption and deaths from heart disease.
    A table lists data for the relation between wine consumption and deaths from heart disease.

    Source: New York Times

    1. Use the statistical menu of your graphing utility to enter the 19 ordered pairs of data items shown in the table.

    2. Use the scatter plot capability to draw a scatter plot of the data.

    3. Select the linear regression option. Use your utility to obtain values for a and b for the equation of the regression line, y=ax+b. You may also be given a correlation coefficient, r. Values of r close to 1 indicate that the points can be described by a linear relationship and the regression line has a positive slope. Values of r close to 1 indicate that the points can be described by a linear relationship and the regression line has a negative slope. Values of r close to 0 indicate no linear relationship between the variables. In this case, a linear model does not accurately describe the data.

    4. Use the appropriate sequence (consult your manual) to graph the regression equation on top of the points in the scatter plot.

Critical Thinking Exercises

Make Sense? In Exercises 105108, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 105. The graph of my linear function at first increased, reached a maximum point, and then decreased.

  2. 106. A linear function that models tuition and fees at public four-year colleges from 2000 through 2020 has negative slope.

  3. 107. Because the variable m does not appear in Ax+By+C=0, equations in this form make it impossible to determine the line’s slope.

  4. 108. The federal minimum wage was $7.25 per hour from 2009 through 2020, so f(x)=7.25 models the minimum wage, f(x), in dollars, for the domain {2009, 2010, 2011, , 2020}.

In Exercises 109112, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 109. The equation y=mx+b shows that no line can have a y-intercept that is numerically equal to its slope.

  2. 110. Every line in the rectangular coordinate system has an equation that can be expressed in slope-intercept form.

  3. 111. The graph of the linear function 5x+6y30=0 is a line passing through the point (6, 0) with slope 56.

  4. 112. The graph of x=7 in the rectangular coordinate system is the single point (7, 0).

In Exercises 113114, find the coefficients that must be placed in each shaded area so that the function’s graph will be a line satisfying the specified conditions.

  1. 113.  x+ y12=0;x-intercept=2;y-intercept=4

  2. 114.  x+ y12=0;y-intercept=6;slope=12

  3. 115. Prove that the equation of a line passing through (a, 0) and (0, b)(a0, b0) can be written in the form xa+yb=1. Why is this called the intercept form of a line?

  4. 116. Excited about the success of celebrity stamps, post office officials were rumored to have put forth a plan to institute two new types of thermometers. On these new scales, °U represents degrees Usher and °R represents degrees Rihanna. If it is known that 40°U=25°R, 280°U=125°R, and degrees Usher is linearly related to degrees Rihanna, write an equation expressing U in terms of R.

Group Exercise

  1. 117. In Exercises 8788, we used the data in a bar graph to develop linear functions that modeled trust in government and trust in media. For this group exercise, you might find it helpful to pattern your work after Exercises 87 and 88. Group members should begin by consulting an almanac, newspaper, magazine, or the Internet to find data that appear to lie approximately on or near a line. Working by hand or using a graphing utility, group members should construct scatter plots for the data that were assembled. If working by hand, draw a line that approximately fits the data in each scatter plot and then write its equation as a function in slope-intercept form. If using a graphing utility, obtain the equation of each regression line. Then use each linear function’s equation to make predictions about what might occur in the future. Are there circumstances that might affect the accuracy of the prediction? List some of these circumstances.

Retaining the Concepts

  1. 118. According to the U.S. Office of Management and Budget, the cost of maintaining existing public transportation infrastructure in 2020 was $103.4 billion and is projected to increase by $0.9 billion each year. By which year is the cost of maintaining existing public transportation infrastructure expected to reach $116 billion?

    (Section P.8, Example 2)

In Exercises 119120, solve and graph the solution set on a number line.

  1. 119. x+34  x23+1

    (Section P.9, Example 5)

  2. 120. 3|2x+6|9<15

    (Section P.9, Example 8)

Preview Exercises

Exercises 121123 will help you prepare for the material covered in the next section.

  1. 121. Write the slope-intercept form of the equation of the line passing through (3, 1) whose slope is the same as the line whose equation is y=2x+1.

  2. 122. Write an equation in general form of the line passing through (3, 5) whose slope is the negative reciprocal (the reciprocal with the opposite sign) of 14.

  3. 123. If f(x)=x2, find

    f(x2)f(x1)x2x1,

    where x1=1 and x2=4.

1.4: Exercise Set

1.4 Exercise Set

Practice Exercises

In Exercises 110, find the slope of the line passing through each pair of points or state that the slope is undefined. Then indicate whether the line through the points rises, falls, is horizontal, or is vertical.

  1. 1. (4, 7) and (8, 10)

  2. 2. (2, 1) and (3, 4)

  3. 3. (2, 1) and (2, 2)

  4. 4. (1, 3) and (2, 4)

  5. 5. (4, 2) and (3, 2)

  6. 6. (4, 1) and (3, 1)

  7. 7. (2, 4) and (1, 1)

  8. 8. (6, 4) and (4, 2)

  9. 9. (5, 3) and (5, 2)

  10. 10. (3, 4) and (3, 5)

In Exercises 1138, use the given conditions to write an equation for each line in point-slope form and slope-intercept form.

  1. 11. Slope=2, passing through (3, 5)

  2. 12. Slope=4, passing through (1, 3)

  3. 13. Slope=6, passing through (2, 5)

  4. 14. Slope=8, passing through (4, 1)

  5. 15. Slope=3, passing through (2, 3)

  6. 16. Slope=5, passing through (4, 2)

  7. 17. Slope=4, passing through (4, 0)

  8. 18. Slope=2, passing through (0, 3)

  9. 19. Slope=1, passing through ( 12, 2)

  10. 20. Slope=1, passing through (4, 14)

  11. 21. Slope=12, passing through the origin

  12. 22. Slope=13, passing through the origin

  13. 23. Slope= 23, passing through (6, 2)

  14. 24. Slope= 35, passing through (10, 4)

  15. 25. Passing through (1, 2) and (5, 10)

  16. 26. Passing through (3, 5) and (8, 15)

  17. 27. Passing through (3, 0) and (0, 3)

  18. 28. Passing through (2, 0) and (0, 2)

  19. 29. Passing through (3, 1) and (2, 4)

  20. 30. Passing through (2, 4) and (1, 1)

  21. 31. Passing through (3, 2) and (3, 6)

  22. 32. Passing through (3, 6) and (3, 2)

  23. 33. Passing through (3, 1) and (4, 1)

  24. 34. Passing through (2, 5) and (6, 5)

  25. 35. Passing through (2, 4) with x-intercept=2

  26. 36. Passing through (1, 3) with x-intercept=1

  27. 37. x-intercept= 12 and y-intercept=4

  28. 38. x-intercept=4 and y-intercept=2

In Exercises 3948, give the slope and y-intercept of each line whose equation is given. Then graph the linear function.

  1. 39. y=2x+1

  2. 40. y=3x+2

  3. 41. f(x)=2x+1

  4. 42. f(x)=3x+2

  5. 43. f(x)=34 x2

  6. 44. f(x)=34 x3

  7. 45. y= 35 x+7

  8. 46. y= 25 x+6

  9. 47. g(x)= 12 x

  10. 48. g(x)= 13 x

In Exercises 4958, graph each equation in a rectangular coordinate system.

  1. 49. y=2

  2. 50. y=4

  3. 51. x=3

  4. 52. x=5

  5. 53. y=0

  6. 54. x=0

  7. 55. f(x)=1

  8. 56. f(x)=3

  9. 57. 3x18=0

  10. 58. 3x+12=0

In Exercises 5966,

  1. Rewrite the given equation in slope-intercept form.

  2. Give the slope and y-intercept.

  3. Use the slope and y-intercept to graph the linear function.

  1. 59. 3x+y5=0

  2. 60. 4x+y6=0

  3. 61. 2x+3y18=0

  4. 62. 4x+6y+12=0

  5. 63. 8x4y12=0

  6. 64. 6x5y20=0

  7. 65. 3y9=0

  8. 66. 4y+28=0

In Exercises 6772, use intercepts to graph each equation.

  1. 67. 6x2y12=0

  2. 68. 6x9y18=0

  3. 69. 2x+3y+6=0

  4. 70. 3x+5y+15=0

  5. 71. 8x2y+12=0

  6. 72. 6x3y+15=0

Practice PLUS

In Exercises 7376, find the slope of the line passing through each pair of points or state that the slope is undefined. Assume that all variables represent positive real numbers. Then indicate whether the line through the points rises, falls, is horizontal, or is vertical.

  1. 73. (0, a) and (b, 0)

  2. 74. (a, 0) and (0, b)

  3. 75. (a, b) and (a, b+c)

  4. 76. (ab, c) and (a, a+c)

In Exercises 7778, give the slope and y-intercept of each line whose equation is given. Assume that B0.

  1. 77. Ax+By=C

  2. 78. Ax=ByC

In Exercises 7980, find the value of y if the line through the two given points is to have the indicated slope.

  1. 79. (3, y) and (1, 4), m=3

  2. 80. (2, y) and (4, 4), m=13

In Exercises 8182, graph each linear function.

  1. 81. 3x4f(x)6=0

  2. 82. 6x5f(x)20=0

  3. 83. If one point on a line is (3, 1) and the line’s slope is 2, find the y-intercept.

  4. 84. If one point on a line is (2, 6) and the line’s slope is  32, find the y-intercept.

Use the figure to make the lists in Exercises 8586.

The image plots a graph of four lines.
  1. 85. List the slopes m1, m2, m3, and m4 in order of decreasing size.

  2. 86. List the y-intercepts b1, b2, b3, and b4 in order of decreasing size.

Application Exercises

Americans’ trust in government and the media has generally been on a downward trend since pollsters first asked questions on these topics in the second half of the twentieth century. Trust in government hit an all-time low of 14% in 2014, while trust in the media bottomed out at 32% in 2016. The bar graph shows the percentage of Americans trusting in the government and the media for five selected years. The data are displayed as two sets of five points each, one scatter plot for the percentage of Americans trusting in the government and one for the percentage of Americans trusting in the media. Also shown for each scatter plot is a line that passes through or near the five points. Use these lines to solve Exercises 8788.

The image shows two graphs titled, Americans trust in Government and media.

Sources: Gallup, Pew Research

  1. 87. In this exercise, you will use the red line for trust in government shown on the scatter plot to develop a model for the percentage of Americans trusting in government.

    1. Use the two points whose coordinates are shown by the voice balloons to find the point-slope form of the equation of the line that models the percentage of Americans trusting in government, y, x years after 2003. Round the slope to two decimal places.

    2. Write the equation from part (a) in slope-intercept form. Use function notation.

    3. Use the linear function to predict the percentage of Americans trusting in government in 2025. Round to the nearest percent.

  2. 88. In this exercise, you will use the blue line for trust in media shown on the scatter plot to develop a model for the percentage of Americans trusting in media.

    1. Use the two points whose coordinates are shown by the voice balloons to find the point-slope form of the equation of the line that models the percentage of Americans trusting in media, y, x years after 2003. Round the slope to two decimal places.

    2. Write the equation from part (a) in slope-intercept form. Use function notation.

    3. Use the linear function to predict the percentage of Americans trusting in media in 2025. Round to the nearest percent.

The bar graph gives the life expectancy for American men and women born in six selected years. In Exercises 8990, you will use the data to obtain models for life expectancy and make predictions about how long American men and women will live in the future.

The image shows a graph bar graph titled, life expectancy in the United States, by year of birth.

Source: National Center for Health Statistics

  1. 89. Use the data for males shown in the bar graph at the bottom of the previous column to solve this exercise.

    1. Let x represent the number of birth years after 1960 and let y represent male life expectancy. Create a scatter plot that displays the data as a set of six points in a rectangular coordinate system.

    2. Draw a line through the two points that show male life expectancies for 1980 and 2000. Use the coordinates of these points to write a linear function that models life expectancy, E(x), for American men born x years after 1960.

    3. Use the function from part (b) to project the life expectancy of American men born in 2020.

  2. 90. Use the data for females shown in the bar graph at the bottom of the previous column to solve this exercise.

    1. Let x represent the number of birth years after 1960 and let y represent female life expectancy. Create a scatter plot that displays the data as a set of six points in a rectangular coordinate system.

    2. Draw a line through the two points that show female life expectancies for 1970 and 2000. Use the coordinates of these points to write a linear function that models life expectancy, E(x), for American women born x years after 1960. Round the slope to two decimal places.

    3. Use the function from part (b) to project the life expectancy of American women born in 2020.

  3. 91. Shown, again, is the scatter plot that indicates a relationship between the percentage of adult females in a country who are literate and the mortality of children under five. Also shown is a line that passes through or near the points. Find a linear function that models the data by finding the slope-intercept form of the line’s equation. Use the function to make a prediction about child mortality based on the percentage of adult females in a country who are literate.

    The image shows a scatterplot titled literacy and child mortality.

    Source: United Nations

  4. 92. Just as money doesn’t buy happiness for individuals, the two don’t necessarily go together for countries either. However, the scatter plot does show a relationship between a country’s annual per capita income and the percentage of people in that country who call themselves “happy.”

    The image shows a graph titled, per capita income and national happiness.

    Source: Richard Layard, Happiness: Lessons from a New Science, Penguin, 2005

    Draw a line that fits the data so that the spread of the data points around the line is as small as possible. Use the coordinates of two points along your line to write the slope-intercept form of its equation. Express the equation in function notation and use the linear function to make a prediction about national happiness based on per capita income.

Explaining the Concepts

  1. 93. What is the slope of a line and how is it found?

  2. 94. Describe how to write the equation of a line if the coordinates of two points along the line are known.

  3. 95. Explain how to derive the slope-intercept form of a line’s equation, y=mx+b, from the point-slope form

    yy1=m(xx1).
  4. 96. Explain how to graph the equation x=2. Can this equation be expressed in slope-intercept form? Explain.

  5. 97. Explain how to use the general form of a line’s equation to find the line’s slope and y-intercept.

  6. 98. Explain how to use intercepts to graph the general form of a line’s equation.

  7. 99. Take another look at the scatter plot in Exercise 91. Although there is a relationship between literacy and child mortality, we cannot conclude that increased literacy causes child mortality to decrease. Offer two or more possible explanations for the data in the scatter plot.

Technology Exercises

Use a graphing utility to graph each equation in Exercises 100103. Then use the  TRACE  feature to trace along the line and find the coordinates of two points. Use these points to compute the line’s slope. Check your result by using the coefficient of x in the line’s equation.

  1. 100. y=2x+4

  2. 101. y=3x+6

  3. 102. y= 12 x5

  4. 103. y=34 x2

  5. 104. Is there a relationship between wine consumption and deaths from heart disease? The table gives data from 19 developed countries.

    A table lists data for the relation between wine consumption and deaths from heart disease.
    A table lists data for the relation between wine consumption and deaths from heart disease.

    Source: New York Times

    1. Use the statistical menu of your graphing utility to enter the 19 ordered pairs of data items shown in the table.

    2. Use the scatter plot capability to draw a scatter plot of the data.

    3. Select the linear regression option. Use your utility to obtain values for a and b for the equation of the regression line, y=ax+b. You may also be given a correlation coefficient, r. Values of r close to 1 indicate that the points can be described by a linear relationship and the regression line has a positive slope. Values of r close to 1 indicate that the points can be described by a linear relationship and the regression line has a negative slope. Values of r close to 0 indicate no linear relationship between the variables. In this case, a linear model does not accurately describe the data.

    4. Use the appropriate sequence (consult your manual) to graph the regression equation on top of the points in the scatter plot.

Critical Thinking Exercises

Make Sense? In Exercises 105108, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 105. The graph of my linear function at first increased, reached a maximum point, and then decreased.

  2. 106. A linear function that models tuition and fees at public four-year colleges from 2000 through 2020 has negative slope.

  3. 107. Because the variable m does not appear in Ax+By+C=0, equations in this form make it impossible to determine the line’s slope.

  4. 108. The federal minimum wage was $7.25 per hour from 2009 through 2020, so f(x)=7.25 models the minimum wage, f(x), in dollars, for the domain {2009, 2010, 2011, , 2020}.

In Exercises 109112, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 109. The equation y=mx+b shows that no line can have a y-intercept that is numerically equal to its slope.

  2. 110. Every line in the rectangular coordinate system has an equation that can be expressed in slope-intercept form.

  3. 111. The graph of the linear function 5x+6y30=0 is a line passing through the point (6, 0) with slope 56.

  4. 112. The graph of x=7 in the rectangular coordinate system is the single point (7, 0).

In Exercises 113114, find the coefficients that must be placed in each shaded area so that the function’s graph will be a line satisfying the specified conditions.

  1. 113.  x+ y12=0;x-intercept=2;y-intercept=4

  2. 114.  x+ y12=0;y-intercept=6;slope=12

  3. 115. Prove that the equation of a line passing through (a, 0) and (0, b)(a0, b0) can be written in the form xa+yb=1. Why is this called the intercept form of a line?

  4. 116. Excited about the success of celebrity stamps, post office officials were rumored to have put forth a plan to institute two new types of thermometers. On these new scales, °U represents degrees Usher and °R represents degrees Rihanna. If it is known that 40°U=25°R, 280°U=125°R, and degrees Usher is linearly related to degrees Rihanna, write an equation expressing U in terms of R.

Group Exercise

  1. 117. In Exercises 8788, we used the data in a bar graph to develop linear functions that modeled trust in government and trust in media. For this group exercise, you might find it helpful to pattern your work after Exercises 87 and 88. Group members should begin by consulting an almanac, newspaper, magazine, or the Internet to find data that appear to lie approximately on or near a line. Working by hand or using a graphing utility, group members should construct scatter plots for the data that were assembled. If working by hand, draw a line that approximately fits the data in each scatter plot and then write its equation as a function in slope-intercept form. If using a graphing utility, obtain the equation of each regression line. Then use each linear function’s equation to make predictions about what might occur in the future. Are there circumstances that might affect the accuracy of the prediction? List some of these circumstances.

Retaining the Concepts

  1. 118. According to the U.S. Office of Management and Budget, the cost of maintaining existing public transportation infrastructure in 2020 was $103.4 billion and is projected to increase by $0.9 billion each year. By which year is the cost of maintaining existing public transportation infrastructure expected to reach $116 billion?

    (Section P.8, Example 2)

In Exercises 119120, solve and graph the solution set on a number line.

  1. 119. x+34  x23+1

    (Section P.9, Example 5)

  2. 120. 3|2x+6|9<15

    (Section P.9, Example 8)

Preview Exercises

Exercises 121123 will help you prepare for the material covered in the next section.

  1. 121. Write the slope-intercept form of the equation of the line passing through (3, 1) whose slope is the same as the line whose equation is y=2x+1.

  2. 122. Write an equation in general form of the line passing through (3, 5) whose slope is the negative reciprocal (the reciprocal with the opposite sign) of 14.

  3. 123. If f(x)=x2, find

    f(x2)f(x1)x2x1,

    where x1=1 and x2=4.

1.4: Exercise Set

1.4 Exercise Set

Practice Exercises

In Exercises 110, find the slope of the line passing through each pair of points or state that the slope is undefined. Then indicate whether the line through the points rises, falls, is horizontal, or is vertical.

  1. 1. (4, 7) and (8, 10)

  2. 2. (2, 1) and (3, 4)

  3. 3. (2, 1) and (2, 2)

  4. 4. (1, 3) and (2, 4)

  5. 5. (4, 2) and (3, 2)

  6. 6. (4, 1) and (3, 1)

  7. 7. (2, 4) and (1, 1)

  8. 8. (6, 4) and (4, 2)

  9. 9. (5, 3) and (5, 2)

  10. 10. (3, 4) and (3, 5)

In Exercises 1138, use the given conditions to write an equation for each line in point-slope form and slope-intercept form.

  1. 11. Slope=2, passing through (3, 5)

  2. 12. Slope=4, passing through (1, 3)

  3. 13. Slope=6, passing through (2, 5)

  4. 14. Slope=8, passing through (4, 1)

  5. 15. Slope=3, passing through (2, 3)

  6. 16. Slope=5, passing through (4, 2)

  7. 17. Slope=4, passing through (4, 0)

  8. 18. Slope=2, passing through (0, 3)

  9. 19. Slope=1, passing through ( 12, 2)

  10. 20. Slope=1, passing through (4, 14)

  11. 21. Slope=12, passing through the origin

  12. 22. Slope=13, passing through the origin

  13. 23. Slope= 23, passing through (6, 2)

  14. 24. Slope= 35, passing through (10, 4)

  15. 25. Passing through (1, 2) and (5, 10)

  16. 26. Passing through (3, 5) and (8, 15)

  17. 27. Passing through (3, 0) and (0, 3)

  18. 28. Passing through (2, 0) and (0, 2)

  19. 29. Passing through (3, 1) and (2, 4)

  20. 30. Passing through (2, 4) and (1, 1)

  21. 31. Passing through (3, 2) and (3, 6)

  22. 32. Passing through (3, 6) and (3, 2)

  23. 33. Passing through (3, 1) and (4, 1)

  24. 34. Passing through (2, 5) and (6, 5)

  25. 35. Passing through (2, 4) with x-intercept=2

  26. 36. Passing through (1, 3) with x-intercept=1

  27. 37. x-intercept= 12 and y-intercept=4

  28. 38. x-intercept=4 and y-intercept=2

In Exercises 3948, give the slope and y-intercept of each line whose equation is given. Then graph the linear function.

  1. 39. y=2x+1

  2. 40. y=3x+2

  3. 41. f(x)=2x+1

  4. 42. f(x)=3x+2

  5. 43. f(x)=34 x2

  6. 44. f(x)=34 x3

  7. 45. y= 35 x+7

  8. 46. y= 25 x+6

  9. 47. g(x)= 12 x

  10. 48. g(x)= 13 x

In Exercises 4958, graph each equation in a rectangular coordinate system.

  1. 49. y=2

  2. 50. y=4

  3. 51. x=3

  4. 52. x=5

  5. 53. y=0

  6. 54. x=0

  7. 55. f(x)=1

  8. 56. f(x)=3

  9. 57. 3x18=0

  10. 58. 3x+12=0

In Exercises 5966,

  1. Rewrite the given equation in slope-intercept form.

  2. Give the slope and y-intercept.

  3. Use the slope and y-intercept to graph the linear function.

  1. 59. 3x+y5=0

  2. 60. 4x+y6=0

  3. 61. 2x+3y18=0

  4. 62. 4x+6y+12=0

  5. 63. 8x4y12=0

  6. 64. 6x5y20=0

  7. 65. 3y9=0

  8. 66. 4y+28=0

In Exercises 6772, use intercepts to graph each equation.

  1. 67. 6x2y12=0

  2. 68. 6x9y18=0

  3. 69. 2x+3y+6=0

  4. 70. 3x+5y+15=0

  5. 71. 8x2y+12=0

  6. 72. 6x3y+15=0

Practice PLUS

In Exercises 7376, find the slope of the line passing through each pair of points or state that the slope is undefined. Assume that all variables represent positive real numbers. Then indicate whether the line through the points rises, falls, is horizontal, or is vertical.

  1. 73. (0, a) and (b, 0)

  2. 74. (a, 0) and (0, b)

  3. 75. (a, b) and (a, b+c)

  4. 76. (ab, c) and (a, a+c)

In Exercises 7778, give the slope and y-intercept of each line whose equation is given. Assume that B0.

  1. 77. Ax+By=C

  2. 78. Ax=ByC

In Exercises 7980, find the value of y if the line through the two given points is to have the indicated slope.

  1. 79. (3, y) and (1, 4), m=3

  2. 80. (2, y) and (4, 4), m=13

In Exercises 8182, graph each linear function.

  1. 81. 3x4f(x)6=0

  2. 82. 6x5f(x)20=0

  3. 83. If one point on a line is (3, 1) and the line’s slope is 2, find the y-intercept.

  4. 84. If one point on a line is (2, 6) and the line’s slope is  32, find the y-intercept.

Use the figure to make the lists in Exercises 8586.

The image plots a graph of four lines.
  1. 85. List the slopes m1, m2, m3, and m4 in order of decreasing size.

  2. 86. List the y-intercepts b1, b2, b3, and b4 in order of decreasing size.

Application Exercises

Americans’ trust in government and the media has generally been on a downward trend since pollsters first asked questions on these topics in the second half of the twentieth century. Trust in government hit an all-time low of 14% in 2014, while trust in the media bottomed out at 32% in 2016. The bar graph shows the percentage of Americans trusting in the government and the media for five selected years. The data are displayed as two sets of five points each, one scatter plot for the percentage of Americans trusting in the government and one for the percentage of Americans trusting in the media. Also shown for each scatter plot is a line that passes through or near the five points. Use these lines to solve Exercises 8788.

The image shows two graphs titled, Americans trust in Government and media.

Sources: Gallup, Pew Research

  1. 87. In this exercise, you will use the red line for trust in government shown on the scatter plot to develop a model for the percentage of Americans trusting in government.

    1. Use the two points whose coordinates are shown by the voice balloons to find the point-slope form of the equation of the line that models the percentage of Americans trusting in government, y, x years after 2003. Round the slope to two decimal places.

    2. Write the equation from part (a) in slope-intercept form. Use function notation.

    3. Use the linear function to predict the percentage of Americans trusting in government in 2025. Round to the nearest percent.

  2. 88. In this exercise, you will use the blue line for trust in media shown on the scatter plot to develop a model for the percentage of Americans trusting in media.

    1. Use the two points whose coordinates are shown by the voice balloons to find the point-slope form of the equation of the line that models the percentage of Americans trusting in media, y, x years after 2003. Round the slope to two decimal places.

    2. Write the equation from part (a) in slope-intercept form. Use function notation.

    3. Use the linear function to predict the percentage of Americans trusting in media in 2025. Round to the nearest percent.

The bar graph gives the life expectancy for American men and women born in six selected years. In Exercises 8990, you will use the data to obtain models for life expectancy and make predictions about how long American men and women will live in the future.

The image shows a graph bar graph titled, life expectancy in the United States, by year of birth.

Source: National Center for Health Statistics

  1. 89. Use the data for males shown in the bar graph at the bottom of the previous column to solve this exercise.

    1. Let x represent the number of birth years after 1960 and let y represent male life expectancy. Create a scatter plot that displays the data as a set of six points in a rectangular coordinate system.

    2. Draw a line through the two points that show male life expectancies for 1980 and 2000. Use the coordinates of these points to write a linear function that models life expectancy, E(x), for American men born x years after 1960.

    3. Use the function from part (b) to project the life expectancy of American men born in 2020.

  2. 90. Use the data for females shown in the bar graph at the bottom of the previous column to solve this exercise.

    1. Let x represent the number of birth years after 1960 and let y represent female life expectancy. Create a scatter plot that displays the data as a set of six points in a rectangular coordinate system.

    2. Draw a line through the two points that show female life expectancies for 1970 and 2000. Use the coordinates of these points to write a linear function that models life expectancy, E(x), for American women born x years after 1960. Round the slope to two decimal places.

    3. Use the function from part (b) to project the life expectancy of American women born in 2020.

  3. 91. Shown, again, is the scatter plot that indicates a relationship between the percentage of adult females in a country who are literate and the mortality of children under five. Also shown is a line that passes through or near the points. Find a linear function that models the data by finding the slope-intercept form of the line’s equation. Use the function to make a prediction about child mortality based on the percentage of adult females in a country who are literate.

    The image shows a scatterplot titled literacy and child mortality.

    Source: United Nations

  4. 92. Just as money doesn’t buy happiness for individuals, the two don’t necessarily go together for countries either. However, the scatter plot does show a relationship between a country’s annual per capita income and the percentage of people in that country who call themselves “happy.”

    The image shows a graph titled, per capita income and national happiness.

    Source: Richard Layard, Happiness: Lessons from a New Science, Penguin, 2005

    Draw a line that fits the data so that the spread of the data points around the line is as small as possible. Use the coordinates of two points along your line to write the slope-intercept form of its equation. Express the equation in function notation and use the linear function to make a prediction about national happiness based on per capita income.

Explaining the Concepts

  1. 93. What is the slope of a line and how is it found?

  2. 94. Describe how to write the equation of a line if the coordinates of two points along the line are known.

  3. 95. Explain how to derive the slope-intercept form of a line’s equation, y=mx+b, from the point-slope form

    yy1=m(xx1).
  4. 96. Explain how to graph the equation x=2. Can this equation be expressed in slope-intercept form? Explain.

  5. 97. Explain how to use the general form of a line’s equation to find the line’s slope and y-intercept.

  6. 98. Explain how to use intercepts to graph the general form of a line’s equation.

  7. 99. Take another look at the scatter plot in Exercise 91. Although there is a relationship between literacy and child mortality, we cannot conclude that increased literacy causes child mortality to decrease. Offer two or more possible explanations for the data in the scatter plot.

Technology Exercises

Use a graphing utility to graph each equation in Exercises 100103. Then use the  TRACE  feature to trace along the line and find the coordinates of two points. Use these points to compute the line’s slope. Check your result by using the coefficient of x in the line’s equation.

  1. 100. y=2x+4

  2. 101. y=3x+6

  3. 102. y= 12 x5

  4. 103. y=34 x2

  5. 104. Is there a relationship between wine consumption and deaths from heart disease? The table gives data from 19 developed countries.

    A table lists data for the relation between wine consumption and deaths from heart disease.
    A table lists data for the relation between wine consumption and deaths from heart disease.

    Source: New York Times

    1. Use the statistical menu of your graphing utility to enter the 19 ordered pairs of data items shown in the table.

    2. Use the scatter plot capability to draw a scatter plot of the data.

    3. Select the linear regression option. Use your utility to obtain values for a and b for the equation of the regression line, y=ax+b. You may also be given a correlation coefficient, r. Values of r close to 1 indicate that the points can be described by a linear relationship and the regression line has a positive slope. Values of r close to 1 indicate that the points can be described by a linear relationship and the regression line has a negative slope. Values of r close to 0 indicate no linear relationship between the variables. In this case, a linear model does not accurately describe the data.

    4. Use the appropriate sequence (consult your manual) to graph the regression equation on top of the points in the scatter plot.

Critical Thinking Exercises

Make Sense? In Exercises 105108, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 105. The graph of my linear function at first increased, reached a maximum point, and then decreased.

  2. 106. A linear function that models tuition and fees at public four-year colleges from 2000 through 2020 has negative slope.

  3. 107. Because the variable m does not appear in Ax+By+C=0, equations in this form make it impossible to determine the line’s slope.

  4. 108. The federal minimum wage was $7.25 per hour from 2009 through 2020, so f(x)=7.25 models the minimum wage, f(x), in dollars, for the domain {2009, 2010, 2011, , 2020}.

In Exercises 109112, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 109. The equation y=mx+b shows that no line can have a y-intercept that is numerically equal to its slope.

  2. 110. Every line in the rectangular coordinate system has an equation that can be expressed in slope-intercept form.

  3. 111. The graph of the linear function 5x+6y30=0 is a line passing through the point (6, 0) with slope 56.

  4. 112. The graph of x=7 in the rectangular coordinate system is the single point (7, 0).

In Exercises 113114, find the coefficients that must be placed in each shaded area so that the function’s graph will be a line satisfying the specified conditions.

  1. 113.  x+ y12=0;x-intercept=2;y-intercept=4

  2. 114.  x+ y12=0;y-intercept=6;slope=12

  3. 115. Prove that the equation of a line passing through (a, 0) and (0, b)(a0, b0) can be written in the form xa+yb=1. Why is this called the intercept form of a line?

  4. 116. Excited about the success of celebrity stamps, post office officials were rumored to have put forth a plan to institute two new types of thermometers. On these new scales, °U represents degrees Usher and °R represents degrees Rihanna. If it is known that 40°U=25°R, 280°U=125°R, and degrees Usher is linearly related to degrees Rihanna, write an equation expressing U in terms of R.

Group Exercise

  1. 117. In Exercises 8788, we used the data in a bar graph to develop linear functions that modeled trust in government and trust in media. For this group exercise, you might find it helpful to pattern your work after Exercises 87 and 88. Group members should begin by consulting an almanac, newspaper, magazine, or the Internet to find data that appear to lie approximately on or near a line. Working by hand or using a graphing utility, group members should construct scatter plots for the data that were assembled. If working by hand, draw a line that approximately fits the data in each scatter plot and then write its equation as a function in slope-intercept form. If using a graphing utility, obtain the equation of each regression line. Then use each linear function’s equation to make predictions about what might occur in the future. Are there circumstances that might affect the accuracy of the prediction? List some of these circumstances.

Retaining the Concepts

  1. 118. According to the U.S. Office of Management and Budget, the cost of maintaining existing public transportation infrastructure in 2020 was $103.4 billion and is projected to increase by $0.9 billion each year. By which year is the cost of maintaining existing public transportation infrastructure expected to reach $116 billion?

    (Section P.8, Example 2)

In Exercises 119120, solve and graph the solution set on a number line.

  1. 119. x+34  x23+1

    (Section P.9, Example 5)

  2. 120. 3|2x+6|9<15

    (Section P.9, Example 8)

Preview Exercises

Exercises 121123 will help you prepare for the material covered in the next section.

  1. 121. Write the slope-intercept form of the equation of the line passing through (3, 1) whose slope is the same as the line whose equation is y=2x+1.

  2. 122. Write an equation in general form of the line passing through (3, 5) whose slope is the negative reciprocal (the reciprocal with the opposite sign) of 14.

  3. 123. If f(x)=x2, find

    f(x2)f(x1)x2x1,

    where x1=1 and x2=4.

Section 1.5: More on Slope

Section 1.5 More on Slope

Learning Objectives

What You’ll Learn

  1. 1 Find slopes and equations of parallel and perpendicular lines.

  2. 2 Interpret slope as rate of change.

  3. 3 Find a function’s average rate of change.

As housing prices skyrocket, fewer U.S. young adults are buying homes and more are living with their parents. Figure 1.51 shows that in 2017, 38.4% of 25-to-34-year-old young adults owned homes, a decrease from the percentage displayed for 2000, and 22.0% lived with parents, nearly doubling the 2000 percentage.

Figure 1.51

A graph plots the living arrangements of U S young adults aged from 25 to 34.

Source: urban.org

Figure 1.51 Full Alternative Text

Take a second look at Figure 1.51. The red graph is going down from left to right, indicating a negative rate of change in home ownership among young adults. The green graph is going up from left to right, indicating a positive rate of change in young adults living with parents. In this section, you will learn how to interpret slope as a rate of change. You will also explore the relationships between parallel and perpendicular lines.

Parallel and Perpendicular Lines

Two nonintersecting lines that lie in the same plane are parallel. If two lines do not intersect, the ratio of the vertical change to the horizontal change is the same for both lines. Because two parallel lines have the same “steepness,” they must have the same slope.

Slope and Parallel Lines

  1. If two nonvertical lines are parallel, then they have the same slope.

  2. If two distinct nonvertical lines have the same slope, then they are parallel.

  3. Two distinct vertical lines, both with undefined slopes, are parallel.

Objective 1: Find slopes and equations of parallel and perpendicular lines

  1. Objective 1 Find slopes and equations of parallel and perpendicular lines.

Watch Video

Example 1 Writing Equations of a Line Parallel to a Given Line

Write an equation of the line passing through (3, 1) and parallel to the line whose equation is y=2x+1. Express the equation in point-slope form and slope-intercept form.

Solution

The situation is illustrated in Figure 1.52. We are looking for the equation of the red line passing through (3, 1) and parallel to the blue line whose equation is y=2x+1. How do we obtain the equation of this red line? Notice that the line passes through the point (3, 1). Using the point-slope form of the line’s equation, we have x1=3 and y1=1.

y minus y sub 1 = m left parenthesis x minus x sub 1 right parenthesis, where y sub 1 = 1 and x sub 1 = negative 3.

Figure 1.52

The image shows a graph that plots two parallel lines.

With (x1, y1)=(3, 1), the only thing missing from the equation of the red line is m, the slope. Do we know anything about the slope of either line in Figure 1.52? The answer is yes; we know the slope of the blue line on the right, whose equation is given.

y = 2 x + 1. 2 is labeled, the slope of the blue line on the right in figure 1.52 is 2.

Parallel lines have the same slope. Because the slope of the blue line is 2, the slope of the red line, the line whose equation we must write, is also 2: m=2. We now have values for x1, y1, and m for the red line.

y minus y sub 1 = m left parenthesis x minus x sub 1 right parenthesis, where y sub 1 = 1, m = 2, and x sub 1 = negative 3.

The point-slope form of the red line’s equation is

y1=2[x(3)] or  y1=2(x+3).

Solving for y, we obtain the slope-intercept form of the equation.

y1=2x+6Apply the distributive property.y=2x+7Add 1 to both sides. This is the slope-interceptform,y=mx+b, of the equation. Using functionnotation, the equation is f(x)=2x+7.

Check Point 1

  • Write an equation of the line passing through (2, 5) and parallel to the line whose equation is y=3x+1. Express the equation in point-slope form and slope-intercept form.

Two lines that intersect at a right angle (90°) are said to be perpendicular, shown in Figure 1.53. The relationship between the slopes of perpendicular lines is not as obvious as the relationship between parallel lines. Figure 1.53 shows line AB, with slope cd. Rotate line AB counterclockwise 90° to the left to obtain line AB, perpendicular to line AB. The figure indicates that the rise and the run of the new line are reversed from the original line, but the former rise, the new run, is now negative. This means that the slope of the new line is dc. Notice that the product of the slopes of the two perpendicular lines is 1:

(cd)(dc)=1.

Figure 1.53 Slopes of perpendicular lines

The image shows a graph that plots two lines that are perpendicular to each other. The lines are drawn in the first quadrant.
Figure 1.53 Full Alternative Text

This relationship holds for all perpendicular lines and is summarized in the box at the top of the next page.

Slope and Perpendicular Lines

  1. If two nonvertical lines are perpendicular, then the product of their slopes is 1.

  2. If the product of the slopes of two lines is 1, then the lines are perpendicular.

  3. A horizontal line having zero slope is perpendicular to a vertical line having undefined slope.

An equivalent way of stating this relationship is to say that one line is perpendicular to another line if its slope is the negative reciprocal of the slope of the other line. For example, if a line has slope 5, any line having slope 15 is perpendicular to it. Similarly, if a line has slope 34, any line having slope 43 is perpendicular to it.

Example 2 Writing Equations of a Line Perpendicular to a Given Line

  1. Find the slope of any line that is perpendicular to the line whose equation is x+4y8=0.

  2. Write the equation of the line passing through (3, 5) and perpendicular to the line whose equation is x+4y8=0. Express the equation in general form.

Solution

  1. We begin by writing the equation of the given line, x+4y8=0, in slope-intercept form. Solve for y.

    y = negative 1 fourth x + 2, where negative 1 fourth is labeled, the slope is negative 1 fourth.

    The given line has slope 14. Any line perpendicular to this line has a slope that is the negative reciprocal of 14. Thus, the slope of any perpendicular line is 4.

  2. Let’s begin by writing the point-slope form of the perpendicular line’s equation. Because the line passes through the point (3, 5), we have x1=3 and y1=5. In part (a), we determined that the slope of any line perpendicular to x+4y8=0 is 4, so the slope of this particular perpendicular line must also be 4: m=4.

y minus y sub 1 = m left parenthesis x minus x sub 1 right parenthesis, where y sub 1 = negative 5, m = 4, and x sub 1 = 3.

The point-slope form of the perpendicular line’s equation is

y(5)=4(x3) or  222..y+5=4(x3).

How can we express this equation, y+5=4(x3), in general form (Ax+By+C=0)? We need to obtain zero on one side of the equation. Let’s do this and keep A, the coefficient of x, positive.

y+5=4(x3)This is the point-slope form of theline’s equation.y+5=4x12Apply the distributive property.yy+55=4xy125To obtain 0 on the left, subtract  y  andsubtract 5 on both sides.0=4xy17Simplify.

In general form, the equation of the perpendicular line is 4xy17=0.

Check Point 2

    1. Find the slope of any line that is perpendicular to the line whose equation is x+3y12=0.

    2. Write the equation of the line passing through (2, 6) and perpendicular to the line whose equation is x+3y12=0. Express the equation in general form.

Objective 1: Find slopes and equations of parallel and perpendicular lines

  1. Objective 1 Find slopes and equations of parallel and perpendicular lines.

Watch Video

Example 1 Writing Equations of a Line Parallel to a Given Line

Write an equation of the line passing through (3, 1) and parallel to the line whose equation is y=2x+1. Express the equation in point-slope form and slope-intercept form.

Solution

The situation is illustrated in Figure 1.52. We are looking for the equation of the red line passing through (3, 1) and parallel to the blue line whose equation is y=2x+1. How do we obtain the equation of this red line? Notice that the line passes through the point (3, 1). Using the point-slope form of the line’s equation, we have x1=3 and y1=1.

y minus y sub 1 = m left parenthesis x minus x sub 1 right parenthesis, where y sub 1 = 1 and x sub 1 = negative 3.

Figure 1.52

The image shows a graph that plots two parallel lines.

With (x1, y1)=(3, 1), the only thing missing from the equation of the red line is m, the slope. Do we know anything about the slope of either line in Figure 1.52? The answer is yes; we know the slope of the blue line on the right, whose equation is given.

y = 2 x + 1. 2 is labeled, the slope of the blue line on the right in figure 1.52 is 2.

Parallel lines have the same slope. Because the slope of the blue line is 2, the slope of the red line, the line whose equation we must write, is also 2: m=2. We now have values for x1, y1, and m for the red line.

y minus y sub 1 = m left parenthesis x minus x sub 1 right parenthesis, where y sub 1 = 1, m = 2, and x sub 1 = negative 3.

The point-slope form of the red line’s equation is

y1=2[x(3)] or  y1=2(x+3).

Solving for y, we obtain the slope-intercept form of the equation.

y1=2x+6Apply the distributive property.y=2x+7Add 1 to both sides. This is the slope-interceptform,y=mx+b, of the equation. Using functionnotation, the equation is f(x)=2x+7.

Check Point 1

  • Write an equation of the line passing through (2, 5) and parallel to the line whose equation is y=3x+1. Express the equation in point-slope form and slope-intercept form.

Two lines that intersect at a right angle (90°) are said to be perpendicular, shown in Figure 1.53. The relationship between the slopes of perpendicular lines is not as obvious as the relationship between parallel lines. Figure 1.53 shows line AB, with slope cd. Rotate line AB counterclockwise 90° to the left to obtain line AB, perpendicular to line AB. The figure indicates that the rise and the run of the new line are reversed from the original line, but the former rise, the new run, is now negative. This means that the slope of the new line is dc. Notice that the product of the slopes of the two perpendicular lines is 1:

(cd)(dc)=1.

Figure 1.53 Slopes of perpendicular lines

The image shows a graph that plots two lines that are perpendicular to each other. The lines are drawn in the first quadrant.
Figure 1.53 Full Alternative Text

This relationship holds for all perpendicular lines and is summarized in the box at the top of the next page.

Slope and Perpendicular Lines

  1. If two nonvertical lines are perpendicular, then the product of their slopes is 1.

  2. If the product of the slopes of two lines is 1, then the lines are perpendicular.

  3. A horizontal line having zero slope is perpendicular to a vertical line having undefined slope.

An equivalent way of stating this relationship is to say that one line is perpendicular to another line if its slope is the negative reciprocal of the slope of the other line. For example, if a line has slope 5, any line having slope 15 is perpendicular to it. Similarly, if a line has slope 34, any line having slope 43 is perpendicular to it.

Example 2 Writing Equations of a Line Perpendicular to a Given Line

  1. Find the slope of any line that is perpendicular to the line whose equation is x+4y8=0.

  2. Write the equation of the line passing through (3, 5) and perpendicular to the line whose equation is x+4y8=0. Express the equation in general form.

Solution

  1. We begin by writing the equation of the given line, x+4y8=0, in slope-intercept form. Solve for y.

    y = negative 1 fourth x + 2, where negative 1 fourth is labeled, the slope is negative 1 fourth.

    The given line has slope 14. Any line perpendicular to this line has a slope that is the negative reciprocal of 14. Thus, the slope of any perpendicular line is 4.

  2. Let’s begin by writing the point-slope form of the perpendicular line’s equation. Because the line passes through the point (3, 5), we have x1=3 and y1=5. In part (a), we determined that the slope of any line perpendicular to x+4y8=0 is 4, so the slope of this particular perpendicular line must also be 4: m=4.

y minus y sub 1 = m left parenthesis x minus x sub 1 right parenthesis, where y sub 1 = negative 5, m = 4, and x sub 1 = 3.

The point-slope form of the perpendicular line’s equation is

y(5)=4(x3) or  222..y+5=4(x3).

How can we express this equation, y+5=4(x3), in general form (Ax+By+C=0)? We need to obtain zero on one side of the equation. Let’s do this and keep A, the coefficient of x, positive.

y+5=4(x3)This is the point-slope form of theline’s equation.y+5=4x12Apply the distributive property.yy+55=4xy125To obtain 0 on the left, subtract  y  andsubtract 5 on both sides.0=4xy17Simplify.

In general form, the equation of the perpendicular line is 4xy17=0.

Check Point 2

    1. Find the slope of any line that is perpendicular to the line whose equation is x+3y12=0.

    2. Write the equation of the line passing through (2, 6) and perpendicular to the line whose equation is x+3y12=0. Express the equation in general form.

Objective 1: Find slopes and equations of parallel and perpendicular lines

  1. Objective 1 Find slopes and equations of parallel and perpendicular lines.

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Example 1 Writing Equations of a Line Parallel to a Given Line

Write an equation of the line passing through (3, 1) and parallel to the line whose equation is y=2x+1. Express the equation in point-slope form and slope-intercept form.

Solution

The situation is illustrated in Figure 1.52. We are looking for the equation of the red line passing through (3, 1) and parallel to the blue line whose equation is y=2x+1. How do we obtain the equation of this red line? Notice that the line passes through the point (3, 1). Using the point-slope form of the line’s equation, we have x1=3 and y1=1.

y minus y sub 1 = m left parenthesis x minus x sub 1 right parenthesis, where y sub 1 = 1 and x sub 1 = negative 3.

Figure 1.52

The image shows a graph that plots two parallel lines.

With (x1, y1)=(3, 1), the only thing missing from the equation of the red line is m, the slope. Do we know anything about the slope of either line in Figure 1.52? The answer is yes; we know the slope of the blue line on the right, whose equation is given.

y = 2 x + 1. 2 is labeled, the slope of the blue line on the right in figure 1.52 is 2.

Parallel lines have the same slope. Because the slope of the blue line is 2, the slope of the red line, the line whose equation we must write, is also 2: m=2. We now have values for x1, y1, and m for the red line.

y minus y sub 1 = m left parenthesis x minus x sub 1 right parenthesis, where y sub 1 = 1, m = 2, and x sub 1 = negative 3.

The point-slope form of the red line’s equation is

y1=2[x(3)] or  y1=2(x+3).

Solving for y, we obtain the slope-intercept form of the equation.

y1=2x+6Apply the distributive property.y=2x+7Add 1 to both sides. This is the slope-interceptform,y=mx+b, of the equation. Using functionnotation, the equation is f(x)=2x+7.

Check Point 1

  • Write an equation of the line passing through (2, 5) and parallel to the line whose equation is y=3x+1. Express the equation in point-slope form and slope-intercept form.

Two lines that intersect at a right angle (90°) are said to be perpendicular, shown in Figure 1.53. The relationship between the slopes of perpendicular lines is not as obvious as the relationship between parallel lines. Figure 1.53 shows line AB, with slope cd. Rotate line AB counterclockwise 90° to the left to obtain line AB, perpendicular to line AB. The figure indicates that the rise and the run of the new line are reversed from the original line, but the former rise, the new run, is now negative. This means that the slope of the new line is dc. Notice that the product of the slopes of the two perpendicular lines is 1:

(cd)(dc)=1.

Figure 1.53 Slopes of perpendicular lines

The image shows a graph that plots two lines that are perpendicular to each other. The lines are drawn in the first quadrant.
Figure 1.53 Full Alternative Text

This relationship holds for all perpendicular lines and is summarized in the box at the top of the next page.

Slope and Perpendicular Lines

  1. If two nonvertical lines are perpendicular, then the product of their slopes is 1.

  2. If the product of the slopes of two lines is 1, then the lines are perpendicular.

  3. A horizontal line having zero slope is perpendicular to a vertical line having undefined slope.

An equivalent way of stating this relationship is to say that one line is perpendicular to another line if its slope is the negative reciprocal of the slope of the other line. For example, if a line has slope 5, any line having slope 15 is perpendicular to it. Similarly, if a line has slope 34, any line having slope 43 is perpendicular to it.

Example 2 Writing Equations of a Line Perpendicular to a Given Line

  1. Find the slope of any line that is perpendicular to the line whose equation is x+4y8=0.

  2. Write the equation of the line passing through (3, 5) and perpendicular to the line whose equation is x+4y8=0. Express the equation in general form.

Solution

  1. We begin by writing the equation of the given line, x+4y8=0, in slope-intercept form. Solve for y.

    y = negative 1 fourth x + 2, where negative 1 fourth is labeled, the slope is negative 1 fourth.

    The given line has slope 14. Any line perpendicular to this line has a slope that is the negative reciprocal of 14. Thus, the slope of any perpendicular line is 4.

  2. Let’s begin by writing the point-slope form of the perpendicular line’s equation. Because the line passes through the point (3, 5), we have x1=3 and y1=5. In part (a), we determined that the slope of any line perpendicular to x+4y8=0 is 4, so the slope of this particular perpendicular line must also be 4: m=4.

y minus y sub 1 = m left parenthesis x minus x sub 1 right parenthesis, where y sub 1 = negative 5, m = 4, and x sub 1 = 3.

The point-slope form of the perpendicular line’s equation is

y(5)=4(x3) or  222..y+5=4(x3).

How can we express this equation, y+5=4(x3), in general form (Ax+By+C=0)? We need to obtain zero on one side of the equation. Let’s do this and keep A, the coefficient of x, positive.

y+5=4(x3)This is the point-slope form of theline’s equation.y+5=4x12Apply the distributive property.yy+55=4xy125To obtain 0 on the left, subtract  y  andsubtract 5 on both sides.0=4xy17Simplify.

In general form, the equation of the perpendicular line is 4xy17=0.

Check Point 2

    1. Find the slope of any line that is perpendicular to the line whose equation is x+3y12=0.

    2. Write the equation of the line passing through (2, 6) and perpendicular to the line whose equation is x+3y12=0. Express the equation in general form.

Objective 2: Interpret slope as rate of change

Slope as Rate of Change

  1. Objective 2 Interpret slope as rate of change.

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Slope is defined as the ratio of a change in y to a corresponding change in x. It describes how fast y is changing with respect to x. For a linear function, slope may be interpreted as the rate of change of the dependent variable per unit change in the independent variable.

Our next example shows how slope can be interpreted as a rate of change in an applied situation. When calculating slope in applied problems, keep track of the units in the numerator and the denominator.

Example 3 Slope as a Rate of Change

The line graphs for the living arrangements of young adults are shown again in Figure 1.54. Find the slope of the line segment for the percentage of young adults, age 25 to 34, owning a home. Describe what this slope represents.

Figure 1.54

The image shows a graph that plots the living arrangements of U S young adults aged from 25 to 34.

Source: urban.org

Solution

We let x represent a year and y the percentage of young adults owning a home in that year. The two points shown on the line segment for home ownership have the following coordinates:

(20 00 comma 45.4) is labeled, in 2000 comma 45.4% owned homes. (20 17 comma 38.4), in 20 17 comma 38.4% owned homes.

Now we compute the slope:

The image shows the process of calculating the slope.

The slope indicates that the percentage of U.S. young adults, age 25 to 34, owning a home decreased at a rate of approximately 0.41 each year for the period from 2000 to 2017. The rate of change is 0.41% per year.

Check Point 3

  • Use the ordered pairs in Figure 1.54 on the previous page to find the slope of the green line segment for young adults, age 25 to 34, living with parents. Express the slope correct to two decimal places and describe what it represents.

Objective 3: Find a function’s average rate of change

The Average Rate of Change of a Function

  1. Objective 3 Find a function’s average rate of change.

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If the graph of a function is not a straight line, the average rate of change between any two points is the slope of the line containing the two points. This line is called a secant line. For example, Figure 1.55 shows the graph of a particular man’s height, in inches, as a function of his age, in years. Two points on the graph are labeled: (13, 57) and (18, 76). At age 13, this man was 57 inches tall and at age 18, he was 76 inches tall.

Figure 1.55 Height as a function of age

The image shows a graph that plots height in inches versus age.
Figure 1.55 Full Alternative Text

The man’s average growth rate between ages 13 and 18 is the slope of the secant line containing (13, 57) and (18, 76):

m=Change inyChange inx=76571813=195=3 45.

This man’s average rate of change, or average growth rate, from age 13 to age 18 was 345, or 3.8, inches per year.

The Average Rate of Change of a Function

Let (x1, f(x1)) and (x2, f(x2)) be distinct points on the graph of a function f. (See Figure 1.56.) The average rate of change of f from x1 to x2 denoted by ΔyΔx (read “delta y divided by delta x” or “change in y divided by change in x”), is

ΔyΔx=f(x2)f(x1)x2x1.

Figure 1.56

The y = f of x is a smooth curve that rises to the left and falls to the right.

Example 4 Finding the Average Rate of Change

Find the average rate of change of f(x)=x2 from

  1. x1=0 to x2=1

  2. x1=1 to x2=2

  3. x1=2 to x2=0.

Solution

  1. The average rate of change of f(x)=x2 from x1=0 to x2=1 is

    ΔyΔx=f(x2)f(x1)x2x1=f(1)f(0)10=12021=1.

    Figure 1.57(a) shows the secant line of f(x)=x2 from x1=0 to x2=1. The average rate of change is positive and the function is increasing on the interval (0, 1).

    Figure 1.57(a) The secant line of f(x)=x2 from x1=0 to x2=1

    A graph is an upward opening parabola that falls through (negative 2, 4), with its vertex at (0, 0), and then rises through (1, 1). A secant line intersects the parabola at (0, 0) and (1, 1).
  2. The average rate of change of f(x)=x2 from x1=1 to x2=2 is

    ΔyΔx=f(x2)f(x1)x2x1=f(2)f(1)21=22121=3.

    Figure 1.57(b) shows the secant line of f(x)=x2 from x1=1 to x2=2. The average rate of change is positive and the function is increasing on the interval (1, 2). Can you see that the graph rises more steeply on the interval (1, 2) than on (0, 1)? This is because the average rate of change from x1=1 to x2=2 is greater than the average rate of change from x1=0 to x2=1.

    Figure 1.57(b) The secant line of f(x)=x2 from x1=1 to x2=2

    A graph is an upward opening parabola that falls through (negative 2, 4), to its vertex at (0, 0), and then rises through (1, 1) and (2, 4). A secant line intersects the parabola at (1, 1) and (2, 4).
  3. The average rate of change of f(x)=x2 from x1=2 to x2=0 is

    ΔyΔx=f(x2)f(x1)x2x1=f(0)f(2)0(2)=02(2)22=42=2.

    Figure 1.57(c) shows the secant line of f(x)=x2 from x1=2 to x2=0. The average rate of change is negative and the function is decreasing on the interval (2, 0).

Figure 1.57(c) The secant line of f(x)=x2 from x1=2 to x2=0

A graph is an upward opening parabola that falls through (negative 2, 4) and (negative 1, 1), with its vertex at (0, 0), and then rises through (1, 1) and (2, 4). A secant line intersects the parabola at (negative 2, 4) and (0, 0).

Check Point 4

  • Find the average rate of change of f(x)=x3 from

    1. x1=0 to x2=1

    2. x1=1 to x2=2

    3. x1=2 to x2=0.

Suppose we are interested in the average rate of change of f from x1=x to x2=x+h. In this case, the average rate of change is

ΔyΔx=f(x2)f(x1)x2x1=f(x+h)f(x)x+hx=f(x+h)f(x)h.

Do you recognize the last expression? It is the difference quotient that you used in Section 1.3. Thus, the difference quotient gives the average rate of change of a function from x to x+h. In the difference quotient, h is thought of as a number very close to 0. In this way, the average rate of change can be found for a very short interval.

Example 5 Finding the Average Rate of Change

When a person receives a drug injected into a muscle, the concentration of the drug in the body, measured in milligrams per 100 milliliters, is a function of the time elapsed after the injection, measured in hours. Figure 1.58 shows the graph of such a function, where x represents hours after the injection and f(x) is the drug’s concentration at time x. Find the average rate of change in the drug’s concentration between 3 and 7 hours.

Figure 1.58 Concentration of a drug as a function of time

A graph plots drug concentration in the blood in milligrams per 100 milliliters versus time in hours.

Solution

At 3 hours, the drug’s concentration is 0.05 and at 7 hours, the concentration is 0.02. The average rate of change in its concentration between 3 and 7 hours is

ΔyΔx=f(x2)f(x1)x2x1=f(7)f(3)73=0.020.0573=0.034=0.0075.

The average rate of change is 0.0075. This means that the drug’s concentration is decreasing at an average rate of 0.0075 milligram per 100 milliliters per hour.

Check Point 5

  • Use Figure 1.58 to find the average rate of change in the drug’s concentration between 1 hour and 3 hours.

The average velocity of an object is its change in position divided by the change in time between the starting and ending positions. If a function expresses an object’s position in terms of time, the function’s average rate of change describes the object’s average velocity.

Average Velocity of an Object

Suppose that a function expresses an object’s position, s(t), in terms of time, t. The average velocity of the object from t1 to t2 is

ΔsΔt=s(t2)s(t1)t2t1.

Example 6 Finding Average Velocity

The distance, s(t), in feet, traveled by a ball rolling down a ramp is given by the function

s(t)=5t2,

where t is the time, in seconds, after the ball is released. Find the ball’s average velocity from

  1. t1=2 seconds to t2=3 seconds.

  2. t1=2 seconds to t2=2.5 seconds.

  3. t1=2 seconds to t2=2.01 seconds.

Solution

  1. The ball’s average velocity between 2 and 3 seconds is

    ΔsΔt=s(3)s(2)3 sec2 sec=5325221 sec =45 ft20 ft1 sec =25 ft/sec.
  2. The ball’s average velocity between 2 and 2.5 seconds is

    ΔsΔt=s(2.5)s(2)2.5 sec2 sec=5(2.5)25220.5 sec=31.25 ft20 ft0.5 sec=22.5 ft/sec.

  3. The ball’s average velocity between 2 and 2.01 seconds is

    ΔsΔt=s(2.01)s(2)2.01 sec2 sec=5(2.01)25220.01 sec=20.2005 ft20 ft 0.01 sec=20.05 ft/sec.

In Example 6, observe that each calculation begins at 2 seconds and involves shorter and shorter time intervals. In calculus, this procedure leads to the concept of instantaneous, as opposed to average, velocity. Instantaneous velocity is discussed in the introduction to calculus in Chapter 11.

Check Point 6

  • The distance, s(t), in feet, traveled by a ball rolling down a ramp is given by the function

    s(t)=4t2,

    where t is the time, in seconds, after the ball is released. Find the ball’s average velocity from

    1. t1=1 second to t2=2 seconds.

    2. t1=1 second to t2=1.5 seconds.

    3. t1=1 second to t2=1.01 seconds.

Objective 3: Find a function’s average rate of change

The Average Rate of Change of a Function

  1. Objective 3 Find a function’s average rate of change.

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If the graph of a function is not a straight line, the average rate of change between any two points is the slope of the line containing the two points. This line is called a secant line. For example, Figure 1.55 shows the graph of a particular man’s height, in inches, as a function of his age, in years. Two points on the graph are labeled: (13, 57) and (18, 76). At age 13, this man was 57 inches tall and at age 18, he was 76 inches tall.

Figure 1.55 Height as a function of age

The image shows a graph that plots height in inches versus age.
Figure 1.55 Full Alternative Text

The man’s average growth rate between ages 13 and 18 is the slope of the secant line containing (13, 57) and (18, 76):

m=Change inyChange inx=76571813=195=3 45.

This man’s average rate of change, or average growth rate, from age 13 to age 18 was 345, or 3.8, inches per year.

The Average Rate of Change of a Function

Let (x1, f(x1)) and (x2, f(x2)) be distinct points on the graph of a function f. (See Figure 1.56.) The average rate of change of f from x1 to x2 denoted by ΔyΔx (read “delta y divided by delta x” or “change in y divided by change in x”), is

ΔyΔx=f(x2)f(x1)x2x1.

Figure 1.56

The y = f of x is a smooth curve that rises to the left and falls to the right.

Example 4 Finding the Average Rate of Change

Find the average rate of change of f(x)=x2 from

  1. x1=0 to x2=1

  2. x1=1 to x2=2

  3. x1=2 to x2=0.

Solution

  1. The average rate of change of f(x)=x2 from x1=0 to x2=1 is

    ΔyΔx=f(x2)f(x1)x2x1=f(1)f(0)10=12021=1.

    Figure 1.57(a) shows the secant line of f(x)=x2 from x1=0 to x2=1. The average rate of change is positive and the function is increasing on the interval (0, 1).

    Figure 1.57(a) The secant line of f(x)=x2 from x1=0 to x2=1

    A graph is an upward opening parabola that falls through (negative 2, 4), with its vertex at (0, 0), and then rises through (1, 1). A secant line intersects the parabola at (0, 0) and (1, 1).
  2. The average rate of change of f(x)=x2 from x1=1 to x2=2 is

    ΔyΔx=f(x2)f(x1)x2x1=f(2)f(1)21=22121=3.

    Figure 1.57(b) shows the secant line of f(x)=x2 from x1=1 to x2=2. The average rate of change is positive and the function is increasing on the interval (1, 2). Can you see that the graph rises more steeply on the interval (1, 2) than on (0, 1)? This is because the average rate of change from x1=1 to x2=2 is greater than the average rate of change from x1=0 to x2=1.

    Figure 1.57(b) The secant line of f(x)=x2 from x1=1 to x2=2

    A graph is an upward opening parabola that falls through (negative 2, 4), to its vertex at (0, 0), and then rises through (1, 1) and (2, 4). A secant line intersects the parabola at (1, 1) and (2, 4).
  3. The average rate of change of f(x)=x2 from x1=2 to x2=0 is

    ΔyΔx=f(x2)f(x1)x2x1=f(0)f(2)0(2)=02(2)22=42=2.

    Figure 1.57(c) shows the secant line of f(x)=x2 from x1=2 to x2=0. The average rate of change is negative and the function is decreasing on the interval (2, 0).

Figure 1.57(c) The secant line of f(x)=x2 from x1=2 to x2=0

A graph is an upward opening parabola that falls through (negative 2, 4) and (negative 1, 1), with its vertex at (0, 0), and then rises through (1, 1) and (2, 4). A secant line intersects the parabola at (negative 2, 4) and (0, 0).

Check Point 4

  • Find the average rate of change of f(x)=x3 from

    1. x1=0 to x2=1

    2. x1=1 to x2=2

    3. x1=2 to x2=0.

Suppose we are interested in the average rate of change of f from x1=x to x2=x+h. In this case, the average rate of change is

ΔyΔx=f(x2)f(x1)x2x1=f(x+h)f(x)x+hx=f(x+h)f(x)h.

Do you recognize the last expression? It is the difference quotient that you used in Section 1.3. Thus, the difference quotient gives the average rate of change of a function from x to x+h. In the difference quotient, h is thought of as a number very close to 0. In this way, the average rate of change can be found for a very short interval.

Example 5 Finding the Average Rate of Change

When a person receives a drug injected into a muscle, the concentration of the drug in the body, measured in milligrams per 100 milliliters, is a function of the time elapsed after the injection, measured in hours. Figure 1.58 shows the graph of such a function, where x represents hours after the injection and f(x) is the drug’s concentration at time x. Find the average rate of change in the drug’s concentration between 3 and 7 hours.

Figure 1.58 Concentration of a drug as a function of time

A graph plots drug concentration in the blood in milligrams per 100 milliliters versus time in hours.

Solution

At 3 hours, the drug’s concentration is 0.05 and at 7 hours, the concentration is 0.02. The average rate of change in its concentration between 3 and 7 hours is

ΔyΔx=f(x2)f(x1)x2x1=f(7)f(3)73=0.020.0573=0.034=0.0075.

The average rate of change is 0.0075. This means that the drug’s concentration is decreasing at an average rate of 0.0075 milligram per 100 milliliters per hour.

Check Point 5

  • Use Figure 1.58 to find the average rate of change in the drug’s concentration between 1 hour and 3 hours.

The average velocity of an object is its change in position divided by the change in time between the starting and ending positions. If a function expresses an object’s position in terms of time, the function’s average rate of change describes the object’s average velocity.

Average Velocity of an Object

Suppose that a function expresses an object’s position, s(t), in terms of time, t. The average velocity of the object from t1 to t2 is

ΔsΔt=s(t2)s(t1)t2t1.

Example 6 Finding Average Velocity

The distance, s(t), in feet, traveled by a ball rolling down a ramp is given by the function

s(t)=5t2,

where t is the time, in seconds, after the ball is released. Find the ball’s average velocity from

  1. t1=2 seconds to t2=3 seconds.

  2. t1=2 seconds to t2=2.5 seconds.

  3. t1=2 seconds to t2=2.01 seconds.

Solution

  1. The ball’s average velocity between 2 and 3 seconds is

    ΔsΔt=s(3)s(2)3 sec2 sec=5325221 sec =45 ft20 ft1 sec =25 ft/sec.
  2. The ball’s average velocity between 2 and 2.5 seconds is

    ΔsΔt=s(2.5)s(2)2.5 sec2 sec=5(2.5)25220.5 sec=31.25 ft20 ft0.5 sec=22.5 ft/sec.

  3. The ball’s average velocity between 2 and 2.01 seconds is

    ΔsΔt=s(2.01)s(2)2.01 sec2 sec=5(2.01)25220.01 sec=20.2005 ft20 ft 0.01 sec=20.05 ft/sec.

In Example 6, observe that each calculation begins at 2 seconds and involves shorter and shorter time intervals. In calculus, this procedure leads to the concept of instantaneous, as opposed to average, velocity. Instantaneous velocity is discussed in the introduction to calculus in Chapter 11.

Check Point 6

  • The distance, s(t), in feet, traveled by a ball rolling down a ramp is given by the function

    s(t)=4t2,

    where t is the time, in seconds, after the ball is released. Find the ball’s average velocity from

    1. t1=1 second to t2=2 seconds.

    2. t1=1 second to t2=1.5 seconds.

    3. t1=1 second to t2=1.01 seconds.

Objective 3: Find a function’s average rate of change

The Average Rate of Change of a Function

  1. Objective 3 Find a function’s average rate of change.

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If the graph of a function is not a straight line, the average rate of change between any two points is the slope of the line containing the two points. This line is called a secant line. For example, Figure 1.55 shows the graph of a particular man’s height, in inches, as a function of his age, in years. Two points on the graph are labeled: (13, 57) and (18, 76). At age 13, this man was 57 inches tall and at age 18, he was 76 inches tall.

Figure 1.55 Height as a function of age

The image shows a graph that plots height in inches versus age.
Figure 1.55 Full Alternative Text

The man’s average growth rate between ages 13 and 18 is the slope of the secant line containing (13, 57) and (18, 76):

m=Change inyChange inx=76571813=195=3 45.

This man’s average rate of change, or average growth rate, from age 13 to age 18 was 345, or 3.8, inches per year.

The Average Rate of Change of a Function

Let (x1, f(x1)) and (x2, f(x2)) be distinct points on the graph of a function f. (See Figure 1.56.) The average rate of change of f from x1 to x2 denoted by ΔyΔx (read “delta y divided by delta x” or “change in y divided by change in x”), is

ΔyΔx=f(x2)f(x1)x2x1.

Figure 1.56

The y = f of x is a smooth curve that rises to the left and falls to the right.

Example 4 Finding the Average Rate of Change

Find the average rate of change of f(x)=x2 from

  1. x1=0 to x2=1

  2. x1=1 to x2=2

  3. x1=2 to x2=0.

Solution

  1. The average rate of change of f(x)=x2 from x1=0 to x2=1 is

    ΔyΔx=f(x2)f(x1)x2x1=f(1)f(0)10=12021=1.

    Figure 1.57(a) shows the secant line of f(x)=x2 from x1=0 to x2=1. The average rate of change is positive and the function is increasing on the interval (0, 1).

    Figure 1.57(a) The secant line of f(x)=x2 from x1=0 to x2=1

    A graph is an upward opening parabola that falls through (negative 2, 4), with its vertex at (0, 0), and then rises through (1, 1). A secant line intersects the parabola at (0, 0) and (1, 1).
  2. The average rate of change of f(x)=x2 from x1=1 to x2=2 is

    ΔyΔx=f(x2)f(x1)x2x1=f(2)f(1)21=22121=3.

    Figure 1.57(b) shows the secant line of f(x)=x2 from x1=1 to x2=2. The average rate of change is positive and the function is increasing on the interval (1, 2). Can you see that the graph rises more steeply on the interval (1, 2) than on (0, 1)? This is because the average rate of change from x1=1 to x2=2 is greater than the average rate of change from x1=0 to x2=1.

    Figure 1.57(b) The secant line of f(x)=x2 from x1=1 to x2=2

    A graph is an upward opening parabola that falls through (negative 2, 4), to its vertex at (0, 0), and then rises through (1, 1) and (2, 4). A secant line intersects the parabola at (1, 1) and (2, 4).
  3. The average rate of change of f(x)=x2 from x1=2 to x2=0 is

    ΔyΔx=f(x2)f(x1)x2x1=f(0)f(2)0(2)=02(2)22=42=2.

    Figure 1.57(c) shows the secant line of f(x)=x2 from x1=2 to x2=0. The average rate of change is negative and the function is decreasing on the interval (2, 0).

Figure 1.57(c) The secant line of f(x)=x2 from x1=2 to x2=0

A graph is an upward opening parabola that falls through (negative 2, 4) and (negative 1, 1), with its vertex at (0, 0), and then rises through (1, 1) and (2, 4). A secant line intersects the parabola at (negative 2, 4) and (0, 0).

Check Point 4

  • Find the average rate of change of f(x)=x3 from

    1. x1=0 to x2=1

    2. x1=1 to x2=2

    3. x1=2 to x2=0.

Suppose we are interested in the average rate of change of f from x1=x to x2=x+h. In this case, the average rate of change is

ΔyΔx=f(x2)f(x1)x2x1=f(x+h)f(x)x+hx=f(x+h)f(x)h.

Do you recognize the last expression? It is the difference quotient that you used in Section 1.3. Thus, the difference quotient gives the average rate of change of a function from x to x+h. In the difference quotient, h is thought of as a number very close to 0. In this way, the average rate of change can be found for a very short interval.

Example 5 Finding the Average Rate of Change

When a person receives a drug injected into a muscle, the concentration of the drug in the body, measured in milligrams per 100 milliliters, is a function of the time elapsed after the injection, measured in hours. Figure 1.58 shows the graph of such a function, where x represents hours after the injection and f(x) is the drug’s concentration at time x. Find the average rate of change in the drug’s concentration between 3 and 7 hours.

Figure 1.58 Concentration of a drug as a function of time

A graph plots drug concentration in the blood in milligrams per 100 milliliters versus time in hours.

Solution

At 3 hours, the drug’s concentration is 0.05 and at 7 hours, the concentration is 0.02. The average rate of change in its concentration between 3 and 7 hours is

ΔyΔx=f(x2)f(x1)x2x1=f(7)f(3)73=0.020.0573=0.034=0.0075.

The average rate of change is 0.0075. This means that the drug’s concentration is decreasing at an average rate of 0.0075 milligram per 100 milliliters per hour.

Check Point 5

  • Use Figure 1.58 to find the average rate of change in the drug’s concentration between 1 hour and 3 hours.

The average velocity of an object is its change in position divided by the change in time between the starting and ending positions. If a function expresses an object’s position in terms of time, the function’s average rate of change describes the object’s average velocity.

Average Velocity of an Object

Suppose that a function expresses an object’s position, s(t), in terms of time, t. The average velocity of the object from t1 to t2 is

ΔsΔt=s(t2)s(t1)t2t1.

Example 6 Finding Average Velocity

The distance, s(t), in feet, traveled by a ball rolling down a ramp is given by the function

s(t)=5t2,

where t is the time, in seconds, after the ball is released. Find the ball’s average velocity from

  1. t1=2 seconds to t2=3 seconds.

  2. t1=2 seconds to t2=2.5 seconds.

  3. t1=2 seconds to t2=2.01 seconds.

Solution

  1. The ball’s average velocity between 2 and 3 seconds is

    ΔsΔt=s(3)s(2)3 sec2 sec=5325221 sec =45 ft20 ft1 sec =25 ft/sec.
  2. The ball’s average velocity between 2 and 2.5 seconds is

    ΔsΔt=s(2.5)s(2)2.5 sec2 sec=5(2.5)25220.5 sec=31.25 ft20 ft0.5 sec=22.5 ft/sec.

  3. The ball’s average velocity between 2 and 2.01 seconds is

    ΔsΔt=s(2.01)s(2)2.01 sec2 sec=5(2.01)25220.01 sec=20.2005 ft20 ft 0.01 sec=20.05 ft/sec.

In Example 6, observe that each calculation begins at 2 seconds and involves shorter and shorter time intervals. In calculus, this procedure leads to the concept of instantaneous, as opposed to average, velocity. Instantaneous velocity is discussed in the introduction to calculus in Chapter 11.

Check Point 6

  • The distance, s(t), in feet, traveled by a ball rolling down a ramp is given by the function

    s(t)=4t2,

    where t is the time, in seconds, after the ball is released. Find the ball’s average velocity from

    1. t1=1 second to t2=2 seconds.

    2. t1=1 second to t2=1.5 seconds.

    3. t1=1 second to t2=1.01 seconds.

Objective 3: Find a function’s average rate of change

The Average Rate of Change of a Function

  1. Objective 3 Find a function’s average rate of change.

Watch Video

If the graph of a function is not a straight line, the average rate of change between any two points is the slope of the line containing the two points. This line is called a secant line. For example, Figure 1.55 shows the graph of a particular man’s height, in inches, as a function of his age, in years. Two points on the graph are labeled: (13, 57) and (18, 76). At age 13, this man was 57 inches tall and at age 18, he was 76 inches tall.

Figure 1.55 Height as a function of age

The image shows a graph that plots height in inches versus age.
Figure 1.55 Full Alternative Text

The man’s average growth rate between ages 13 and 18 is the slope of the secant line containing (13, 57) and (18, 76):

m=Change inyChange inx=76571813=195=3 45.

This man’s average rate of change, or average growth rate, from age 13 to age 18 was 345, or 3.8, inches per year.

The Average Rate of Change of a Function

Let (x1, f(x1)) and (x2, f(x2)) be distinct points on the graph of a function f. (See Figure 1.56.) The average rate of change of f from x1 to x2 denoted by ΔyΔx (read “delta y divided by delta x” or “change in y divided by change in x”), is

ΔyΔx=f(x2)f(x1)x2x1.

Figure 1.56

The y = f of x is a smooth curve that rises to the left and falls to the right.

Example 4 Finding the Average Rate of Change

Find the average rate of change of f(x)=x2 from

  1. x1=0 to x2=1

  2. x1=1 to x2=2

  3. x1=2 to x2=0.

Solution

  1. The average rate of change of f(x)=x2 from x1=0 to x2=1 is

    ΔyΔx=f(x2)f(x1)x2x1=f(1)f(0)10=12021=1.

    Figure 1.57(a) shows the secant line of f(x)=x2 from x1=0 to x2=1. The average rate of change is positive and the function is increasing on the interval (0, 1).

    Figure 1.57(a) The secant line of f(x)=x2 from x1=0 to x2=1

    A graph is an upward opening parabola that falls through (negative 2, 4), with its vertex at (0, 0), and then rises through (1, 1). A secant line intersects the parabola at (0, 0) and (1, 1).
  2. The average rate of change of f(x)=x2 from x1=1 to x2=2 is

    ΔyΔx=f(x2)f(x1)x2x1=f(2)f(1)21=22121=3.

    Figure 1.57(b) shows the secant line of f(x)=x2 from x1=1 to x2=2. The average rate of change is positive and the function is increasing on the interval (1, 2). Can you see that the graph rises more steeply on the interval (1, 2) than on (0, 1)? This is because the average rate of change from x1=1 to x2=2 is greater than the average rate of change from x1=0 to x2=1.

    Figure 1.57(b) The secant line of f(x)=x2 from x1=1 to x2=2

    A graph is an upward opening parabola that falls through (negative 2, 4), to its vertex at (0, 0), and then rises through (1, 1) and (2, 4). A secant line intersects the parabola at (1, 1) and (2, 4).
  3. The average rate of change of f(x)=x2 from x1=2 to x2=0 is

    ΔyΔx=f(x2)f(x1)x2x1=f(0)f(2)0(2)=02(2)22=42=2.

    Figure 1.57(c) shows the secant line of f(x)=x2 from x1=2 to x2=0. The average rate of change is negative and the function is decreasing on the interval (2, 0).

Figure 1.57(c) The secant line of f(x)=x2 from x1=2 to x2=0

A graph is an upward opening parabola that falls through (negative 2, 4) and (negative 1, 1), with its vertex at (0, 0), and then rises through (1, 1) and (2, 4). A secant line intersects the parabola at (negative 2, 4) and (0, 0).

Check Point 4

  • Find the average rate of change of f(x)=x3 from

    1. x1=0 to x2=1

    2. x1=1 to x2=2

    3. x1=2 to x2=0.

Suppose we are interested in the average rate of change of f from x1=x to x2=x+h. In this case, the average rate of change is

ΔyΔx=f(x2)f(x1)x2x1=f(x+h)f(x)x+hx=f(x+h)f(x)h.

Do you recognize the last expression? It is the difference quotient that you used in Section 1.3. Thus, the difference quotient gives the average rate of change of a function from x to x+h. In the difference quotient, h is thought of as a number very close to 0. In this way, the average rate of change can be found for a very short interval.

Example 5 Finding the Average Rate of Change

When a person receives a drug injected into a muscle, the concentration of the drug in the body, measured in milligrams per 100 milliliters, is a function of the time elapsed after the injection, measured in hours. Figure 1.58 shows the graph of such a function, where x represents hours after the injection and f(x) is the drug’s concentration at time x. Find the average rate of change in the drug’s concentration between 3 and 7 hours.

Figure 1.58 Concentration of a drug as a function of time

A graph plots drug concentration in the blood in milligrams per 100 milliliters versus time in hours.

Solution

At 3 hours, the drug’s concentration is 0.05 and at 7 hours, the concentration is 0.02. The average rate of change in its concentration between 3 and 7 hours is

ΔyΔx=f(x2)f(x1)x2x1=f(7)f(3)73=0.020.0573=0.034=0.0075.

The average rate of change is 0.0075. This means that the drug’s concentration is decreasing at an average rate of 0.0075 milligram per 100 milliliters per hour.

Check Point 5

  • Use Figure 1.58 to find the average rate of change in the drug’s concentration between 1 hour and 3 hours.

The average velocity of an object is its change in position divided by the change in time between the starting and ending positions. If a function expresses an object’s position in terms of time, the function’s average rate of change describes the object’s average velocity.

Average Velocity of an Object

Suppose that a function expresses an object’s position, s(t), in terms of time, t. The average velocity of the object from t1 to t2 is

ΔsΔt=s(t2)s(t1)t2t1.

Example 6 Finding Average Velocity

The distance, s(t), in feet, traveled by a ball rolling down a ramp is given by the function

s(t)=5t2,

where t is the time, in seconds, after the ball is released. Find the ball’s average velocity from

  1. t1=2 seconds to t2=3 seconds.

  2. t1=2 seconds to t2=2.5 seconds.

  3. t1=2 seconds to t2=2.01 seconds.

Solution

  1. The ball’s average velocity between 2 and 3 seconds is

    ΔsΔt=s(3)s(2)3 sec2 sec=5325221 sec =45 ft20 ft1 sec =25 ft/sec.
  2. The ball’s average velocity between 2 and 2.5 seconds is

    ΔsΔt=s(2.5)s(2)2.5 sec2 sec=5(2.5)25220.5 sec=31.25 ft20 ft0.5 sec=22.5 ft/sec.

  3. The ball’s average velocity between 2 and 2.01 seconds is

    ΔsΔt=s(2.01)s(2)2.01 sec2 sec=5(2.01)25220.01 sec=20.2005 ft20 ft 0.01 sec=20.05 ft/sec.

In Example 6, observe that each calculation begins at 2 seconds and involves shorter and shorter time intervals. In calculus, this procedure leads to the concept of instantaneous, as opposed to average, velocity. Instantaneous velocity is discussed in the introduction to calculus in Chapter 11.

Check Point 6

  • The distance, s(t), in feet, traveled by a ball rolling down a ramp is given by the function

    s(t)=4t2,

    where t is the time, in seconds, after the ball is released. Find the ball’s average velocity from

    1. t1=1 second to t2=2 seconds.

    2. t1=1 second to t2=1.5 seconds.

    3. t1=1 second to t2=1.01 seconds.

1.5: Exercise Set

1.5 Exercise Set

Practice Exercises

In Exercises 14, write an equation for line L in point-slope form and slope-intercept form.

  1. 1.

    The image shows a graph that plots two parallel lines.
  2. 2.

    The image shows a graph that plots two parallel lines.
  3. 3.

    The image shows a graph that plots two perpendicular lines.
  4. 4.

    The image shows a graph that plots two perpendicular lines.

In Exercises 58, use the given conditions to write an equation for each line in point-slope form and slope-intercept form.

  1. 5. Passing through (8, 10) and parallel to the line whose equation is y=4x+3

  2. 6. Passing through (2, 7) and parallel to the line whose equation is y=5x+4

  3. 7. Passing through (2, 3) and perpendicular to the line whose equation is y=15 x+6

  4. 8. Passing through (4, 2) and perpendicular to the line whose equation is y=13 x+7

In Exercises 912, use the given conditions to write an equation for each line in point-slope form and general form.

  1. 9. Passing through (2, 2) and parallel to the line whose equation is 2x3y7=0

  2. 10. Passing through (1, 3) and parallel to the line whose equation is 3x2y5=0

  3. 11. Passing through (4, 7) and perpendicular to the line whose equation is x2y3=0

  4. 12. Passing through (5, 9) and perpendicular to the line whose equation is x+7y12=0

In Exercises 1318, find the average rate of change of the function from x1 to x2.

  1. 13. f(x)=3x from x1=0 to x2=5

  2. 14. f(x)=6x from x1=0 to x2=4

  3. 15. f(x)=x2+2x from x1=3 to x2=5

  4. 16. f(x)=x22x from x1=3 to x2=6

  5. 17. f(x)=x from x1=4 to x2=9

  6. 18. f(x)=x from x1=9 to x2=16

In Exercises 1920, suppose that a ball is rolling down a ramp. The distance traveled by the ball is given by the function in each exercise, where t is the time, in seconds, after the ball is released, and s(t) is measured in feet. For each given function, find the ball’s average velocity from

  1. t1=3 to t2=4.

  2. t1=3 to t2=3.5.

  3. t1=3 to t2=3.01.

  4. t1=3 to t2=3.001.

  1. 19. s(t)=10t2

  2. 20. s(t)=12t2

Practice PLUS

In Exercises 2126, write an equation in slope-intercept form of a linear function f whose graph satisfies the given conditions.

  1. 21. The graph of f passes through (1, 5) and is perpendicular to the line whose equation is x=6.

  2. 22. The graph of f passes through (2, 6) and is perpendicular to the line whose equation is x=4.

  3. 23. The graph of f passes through (6, 4) and is perpendicular to the line that has an x-intercept of 2 and a y-intercept of 4.

  4. 24. The graph of f passes through (5, 6) and is perpendicular to the line that has an x-intercept of 3 and a y-intercept of 9.

  5. 25. The graph of f is perpendicular to the line whose equation is 3x2y4=0 and has the same y-intercept as this line.

  6. 26. The graph of f is perpendicular to the line whose equation is 4xy6=0 and has the same y-intercept as this line.

Application Exercises

The bar graph shows that as costs changed over the decades, Americans devoted less of their budget to groceries and more to health care.

A bar graph provides the percentage of total spending in the United States on food and health care.

Source: Time, October 10, 2011

In Exercises 2728, find a linear function in slope-intercept form that models the given description. Each function should model the percentage of total spending, p(x), by Americans x years after 1950.

  1. 27. In 1950, Americans spent 22% of their budget on food. This has decreased at an average rate of approximately 0.25% per year since then.

  2. 28. In 1950, Americans spent 3% of their budget on health care. This has increased at an average rate of approximately 0.22% per year since then.

The stated intent of the 1994 “don’t ask, don’t tell” policy was to reduce the number of LGBTQ discharges from the military. Nearly 14,000 active-duty LGBTQ servicemembers were dismissed under the policy, which officially ended in 2011, after 18 years. The line graph shows the number of discharges under “don’t ask, don’t tell” from 1994 through 2010. Use the data displayed by the graph to solve Exercises 2930.

A line graph plots the number of active duty L G B T Q service members discharged from the military.

Source: General Accountability Office

(In Exercises 2930, be sure to refer to the graph at the bottom of the previous page.)

  1. 29. Find the average rate of change, rounded to the nearest whole number, from 1994 through 1998. Describe what this means.

  2. 30. Find the average rate of change, rounded to the nearest whole number, from 2001 through 2006. Describe what this means.

The function f(x)=1.1x335x2+264x+557 models the number of discharges, f(x), under “don’t ask, don’t tell” x years after 1994. Use this model and its graph, shown below, to solve Exercises 3132.

The image shows a graph of a model for discharges under do not ask, do not tell.
  1. 31.

    1. Find the slope of the secant line, rounded to the nearest whole number, from x1=0 to x2=4.

    2. Does the slope from part (a) underestimate or overestimate the average yearly increase that you determined in Exercise 29? By how much?

  2. 32.

    1. Find the slope of the secant line, rounded to the nearest whole number, from x1=7 to x2=12.

    2. Does the slope from part (b) underestimate or overestimate the average yearly decrease that you determined in Exercise 30? By how much?

Explaining the Concepts

  1. 33. If two lines are parallel, describe the relationship between their slopes.

  2. 34. If two lines are perpendicular, describe the relationship between their slopes.

  3. 35. If you know a point on a line and you know the equation of a line perpendicular to this line, explain how to write the line’s equation.

  4. 36. A formula in the form y=mx+b models the average retail price, y, of a new car x years after 2000. Would you expect m to be positive, negative, or zero? Explain your answer.

  5. 37. What is a secant line?

  6. 38. What is the average rate of change of a function?

Technology Exercises

  1. 39.

    1. Why are the lines whose equations are y=13 x+1 and y=3x2 perpendicular?

    2. Use a graphing utility to graph the equations in a [10, 10, 1] by [10, 10, 1] viewing rectangle. Do the lines appear to be perpendicular?

    3. Now use the zoom square feature of your utility. Describe what happens to the graphs. Explain why this is so.

Critical Thinking Exercises

Make Sense? In Exercises 4043, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 40. I computed the slope of one line to be 35 and the slope of a second line to be 53, so the lines must be perpendicular.

  2. 41. I have linear functions that model changes for men and women over the same time period. The functions have the same slope, so their graphs are parallel lines, indicating that the rate of change for men is the same as the rate of change for women.

  3. 42. The graph of my function is not a straight line, so I cannot use slope to analyze its rates of change.

  4. 43. According to the Blitzer Bonus here, calculus studies change by analyzing slopes of secant lines over successively shorter intervals.

  5. 44. What is the slope of a line that is perpendicular to the line whose equation is Ax+By+C=0, A0 and B0?

  6. 45. Determine the value of A so that the line whose equation is Ax+y2=0 is perpendicular to the line containing the points (1, 3) and (2, 4).

Retaining the Concepts

  1. 46. Solve and check: 24+3 (x+2)=5(x12). (Section P.7, Example 1)

  2. 47. After a 30% price reduction, you purchase a 50  ˝ 4K UHD TV for $245. What was the television’s price before the reduction? (Section P.8, Example 4)

  3. 48. Solve: x+7+5=x (Section P.7, Example 12)

Preview Exercises

Exercises 4951 will help you prepare for the material covered in the next section. In each exercise, graph the functions in parts (a) and (b) in the same rectangular coordinate system.

  1. 49.

    1. Graph f(x)=|x| using the ordered pairs (3, f(3)), (2, f(2)), (1, f(1)), (0, f(0)), (1, f(1)), (2, f(2)), and (3, f(3)).

    2. Subtract 4 from each y-coordinate of the ordered pairs in part (a). Then graph the ordered pairs and connect them with two linear pieces.

    3. Describe the relationship between the graph in part (b) and the graph in part (a).

  2. 50.

    1. Graph f(x)=x2 using the ordered pairs (3, f(3)), (2, f(2)), (1, f(1)), (0, f(0)), (1, f(1)), (2, f(2)), and (3, f(3)).

    2. Add 2 to each x-coordinate of the ordered pairs in part (a). Then graph the ordered pairs and connect them with a smooth curve.

    3. Describe the relationship between the graph in part (b) and the graph in part (a).

  3. 51.

    1. Graph f(x)=x3 using the ordered pairs (2, f(2)), (1, f(1)), (0, f(0)), (1, f(1)), and (2, f(2)).

    2. Replace each x-coordinate of the ordered pairs in part (a) with its opposite, or additive inverse. Then graph the ordered pairs and connect them with a smooth curve.

    3. Describe the relationship between the graph in part (b) and the graph in part (a).

1.5: Exercise Set

1.5 Exercise Set

Practice Exercises

In Exercises 14, write an equation for line L in point-slope form and slope-intercept form.

  1. 1.

    The image shows a graph that plots two parallel lines.
  2. 2.

    The image shows a graph that plots two parallel lines.
  3. 3.

    The image shows a graph that plots two perpendicular lines.
  4. 4.

    The image shows a graph that plots two perpendicular lines.

In Exercises 58, use the given conditions to write an equation for each line in point-slope form and slope-intercept form.

  1. 5. Passing through (8, 10) and parallel to the line whose equation is y=4x+3

  2. 6. Passing through (2, 7) and parallel to the line whose equation is y=5x+4

  3. 7. Passing through (2, 3) and perpendicular to the line whose equation is y=15 x+6

  4. 8. Passing through (4, 2) and perpendicular to the line whose equation is y=13 x+7

In Exercises 912, use the given conditions to write an equation for each line in point-slope form and general form.

  1. 9. Passing through (2, 2) and parallel to the line whose equation is 2x3y7=0

  2. 10. Passing through (1, 3) and parallel to the line whose equation is 3x2y5=0

  3. 11. Passing through (4, 7) and perpendicular to the line whose equation is x2y3=0

  4. 12. Passing through (5, 9) and perpendicular to the line whose equation is x+7y12=0

In Exercises 1318, find the average rate of change of the function from x1 to x2.

  1. 13. f(x)=3x from x1=0 to x2=5

  2. 14. f(x)=6x from x1=0 to x2=4

  3. 15. f(x)=x2+2x from x1=3 to x2=5

  4. 16. f(x)=x22x from x1=3 to x2=6

  5. 17. f(x)=x from x1=4 to x2=9

  6. 18. f(x)=x from x1=9 to x2=16

In Exercises 1920, suppose that a ball is rolling down a ramp. The distance traveled by the ball is given by the function in each exercise, where t is the time, in seconds, after the ball is released, and s(t) is measured in feet. For each given function, find the ball’s average velocity from

  1. t1=3 to t2=4.

  2. t1=3 to t2=3.5.

  3. t1=3 to t2=3.01.

  4. t1=3 to t2=3.001.

  1. 19. s(t)=10t2

  2. 20. s(t)=12t2

Practice PLUS

In Exercises 2126, write an equation in slope-intercept form of a linear function f whose graph satisfies the given conditions.

  1. 21. The graph of f passes through (1, 5) and is perpendicular to the line whose equation is x=6.

  2. 22. The graph of f passes through (2, 6) and is perpendicular to the line whose equation is x=4.

  3. 23. The graph of f passes through (6, 4) and is perpendicular to the line that has an x-intercept of 2 and a y-intercept of 4.

  4. 24. The graph of f passes through (5, 6) and is perpendicular to the line that has an x-intercept of 3 and a y-intercept of 9.

  5. 25. The graph of f is perpendicular to the line whose equation is 3x2y4=0 and has the same y-intercept as this line.

  6. 26. The graph of f is perpendicular to the line whose equation is 4xy6=0 and has the same y-intercept as this line.

Application Exercises

The bar graph shows that as costs changed over the decades, Americans devoted less of their budget to groceries and more to health care.

A bar graph provides the percentage of total spending in the United States on food and health care.

Source: Time, October 10, 2011

In Exercises 2728, find a linear function in slope-intercept form that models the given description. Each function should model the percentage of total spending, p(x), by Americans x years after 1950.

  1. 27. In 1950, Americans spent 22% of their budget on food. This has decreased at an average rate of approximately 0.25% per year since then.

  2. 28. In 1950, Americans spent 3% of their budget on health care. This has increased at an average rate of approximately 0.22% per year since then.

The stated intent of the 1994 “don’t ask, don’t tell” policy was to reduce the number of LGBTQ discharges from the military. Nearly 14,000 active-duty LGBTQ servicemembers were dismissed under the policy, which officially ended in 2011, after 18 years. The line graph shows the number of discharges under “don’t ask, don’t tell” from 1994 through 2010. Use the data displayed by the graph to solve Exercises 2930.

A line graph plots the number of active duty L G B T Q service members discharged from the military.

Source: General Accountability Office

(In Exercises 2930, be sure to refer to the graph at the bottom of the previous page.)

  1. 29. Find the average rate of change, rounded to the nearest whole number, from 1994 through 1998. Describe what this means.

  2. 30. Find the average rate of change, rounded to the nearest whole number, from 2001 through 2006. Describe what this means.

The function f(x)=1.1x335x2+264x+557 models the number of discharges, f(x), under “don’t ask, don’t tell” x years after 1994. Use this model and its graph, shown below, to solve Exercises 3132.

The image shows a graph of a model for discharges under do not ask, do not tell.
  1. 31.

    1. Find the slope of the secant line, rounded to the nearest whole number, from x1=0 to x2=4.

    2. Does the slope from part (a) underestimate or overestimate the average yearly increase that you determined in Exercise 29? By how much?

  2. 32.

    1. Find the slope of the secant line, rounded to the nearest whole number, from x1=7 to x2=12.

    2. Does the slope from part (b) underestimate or overestimate the average yearly decrease that you determined in Exercise 30? By how much?

Explaining the Concepts

  1. 33. If two lines are parallel, describe the relationship between their slopes.

  2. 34. If two lines are perpendicular, describe the relationship between their slopes.

  3. 35. If you know a point on a line and you know the equation of a line perpendicular to this line, explain how to write the line’s equation.

  4. 36. A formula in the form y=mx+b models the average retail price, y, of a new car x years after 2000. Would you expect m to be positive, negative, or zero? Explain your answer.

  5. 37. What is a secant line?

  6. 38. What is the average rate of change of a function?

Technology Exercises

  1. 39.

    1. Why are the lines whose equations are y=13 x+1 and y=3x2 perpendicular?

    2. Use a graphing utility to graph the equations in a [10, 10, 1] by [10, 10, 1] viewing rectangle. Do the lines appear to be perpendicular?

    3. Now use the zoom square feature of your utility. Describe what happens to the graphs. Explain why this is so.

Critical Thinking Exercises

Make Sense? In Exercises 4043, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 40. I computed the slope of one line to be 35 and the slope of a second line to be 53, so the lines must be perpendicular.

  2. 41. I have linear functions that model changes for men and women over the same time period. The functions have the same slope, so their graphs are parallel lines, indicating that the rate of change for men is the same as the rate of change for women.

  3. 42. The graph of my function is not a straight line, so I cannot use slope to analyze its rates of change.

  4. 43. According to the Blitzer Bonus here, calculus studies change by analyzing slopes of secant lines over successively shorter intervals.

  5. 44. What is the slope of a line that is perpendicular to the line whose equation is Ax+By+C=0, A0 and B0?

  6. 45. Determine the value of A so that the line whose equation is Ax+y2=0 is perpendicular to the line containing the points (1, 3) and (2, 4).

Retaining the Concepts

  1. 46. Solve and check: 24+3 (x+2)=5(x12). (Section P.7, Example 1)

  2. 47. After a 30% price reduction, you purchase a 50  ˝ 4K UHD TV for $245. What was the television’s price before the reduction? (Section P.8, Example 4)

  3. 48. Solve: x+7+5=x (Section P.7, Example 12)

Preview Exercises

Exercises 4951 will help you prepare for the material covered in the next section. In each exercise, graph the functions in parts (a) and (b) in the same rectangular coordinate system.

  1. 49.

    1. Graph f(x)=|x| using the ordered pairs (3, f(3)), (2, f(2)), (1, f(1)), (0, f(0)), (1, f(1)), (2, f(2)), and (3, f(3)).

    2. Subtract 4 from each y-coordinate of the ordered pairs in part (a). Then graph the ordered pairs and connect them with two linear pieces.

    3. Describe the relationship between the graph in part (b) and the graph in part (a).

  2. 50.

    1. Graph f(x)=x2 using the ordered pairs (3, f(3)), (2, f(2)), (1, f(1)), (0, f(0)), (1, f(1)), (2, f(2)), and (3, f(3)).

    2. Add 2 to each x-coordinate of the ordered pairs in part (a). Then graph the ordered pairs and connect them with a smooth curve.

    3. Describe the relationship between the graph in part (b) and the graph in part (a).

  3. 51.

    1. Graph f(x)=x3 using the ordered pairs (2, f(2)), (1, f(1)), (0, f(0)), (1, f(1)), and (2, f(2)).

    2. Replace each x-coordinate of the ordered pairs in part (a) with its opposite, or additive inverse. Then graph the ordered pairs and connect them with a smooth curve.

    3. Describe the relationship between the graph in part (b) and the graph in part (a).

Chapter 1: Mid-Chapter Check Point

Chapter 1 Mid-Chapter Check Point

What You Know: We learned that a function is a relation in which no two ordered pairs have the same first component and different second components. We represented functions as equations and used function notation. We graphed functions and applied the vertical line test to identify graphs of functions. We determined the domain and range of a function from its graph, using inputs on the x-axis for the domain and outputs on the y-axis for the range. We used graphs to identify intervals on which functions increase, decrease, or are constant, as well as to locate relative maxima or minima. We determined when graphs of equations are symmetric with respect to the y-axis (no change when x is substituted for x), the x-axis (no change when y is substituted for y), and the origin (no change when x is substituted for x and y is substituted for y). We identified even functions [f(x)=f(x): y-axis symmetry] and odd functions [f(x)=f(x): origin symmetry]. Finally, we studied linear functions and slope, using slope (change in y divided by change in x) to develop various forms for equations of lines:

Point slope form, y minus y sub 1 = m left parenthesis x minus x sub 1 right parenthesis. Slope intercept form, y = f of x = m x + b. Horizontal line, y = f of x = b. Vertical line, x = a. General form, A x + B y + C = 0.

We saw that parallel lines have the same slope and that perpendicular lines have slopes that are negative reciprocals. For linear functions, slope was interpreted as the rate of change of the dependent variable per unit change in the independent variable. For nonlinear functions, the slope of the secant line between (x1, f(x1)) and (x2, f(x2)) described the average rate of change of f from x1 to x2:ΔyΔx=f(x2)f(x1)x2x1.

In Exercises 16, determine whether each relation is a function. Give the domain and range for each relation.

  1. 1. {(2, 6), (1, 4), (2, 6)}

  2. 2. {(0, 1), (2, 1), (3, 4)}

  3. 3.

    A graph is a line that falls from a closed point at (negative 2, 3) through (0, 1) to (1, 0) and rises to an open point at (2, 1).
  4. 4.

    A graph is a continuous curve that starts at a closed point (negative 3, 0) passes through (0, 2), (0, 0), (0, negative 1), and ends at (4, 0).
  5. 5.

    A graph plots six points as follows, (negative 2, negative 2), (negative 1, negative 2), (negative 1, 1), (0, negative 1), (1, 1), and (2, 3).
  6. 6.

    The graph is a curve that falls through (negative 3, 1) and (0, 0) to a closed point at (1, negative 1). All values are estimated.

In Exercises 78, determine whether each equation defines y as a function of x.

  1. 7. x2+y=5

  2. 8. x+y2=5

Use the graph of f to solve Exercises 924. Where applicable, use interval notation.

The graph of y = f of x is a downward opening parabola that rises through (negative 6, 0) with its vertex at (negative 2, 4) and falls through (2, 0). All values are estimated.
  1. 9. Explain why f represents the graph of a function.

  2. 10. Find the domain of f.

  3. 11. Find the range of f.

  4. 12. Find the x-intercept(s).

  5. 13. Find the y-intercept.

  6. 14. Find the interval(s) on which f is increasing.

  7. 15. Find the interval(s) on which f is decreasing.

  8. 16. At what number does f have a relative maximum?

  9. 17. What is the relative maximum of f?

  10. 18. Find f(4).

  11. 19. For what value or values of x is f(x)=2?

  12. 20. For what value or values of x is f(x)=0?

  13. 21. For what values of x is f(x)>0?

  14. 22. Is f(100) positive or negative?

  15. 23. Is f even, odd, or neither?

  16. 24. Find the average rate of change of f from x1=4 to x2=4.

In Exercises 2526, determine whether the graph of each equation is symmetric with respect to the y-axis, the x-axis, the origin, more than one of these, or none of these.

  1. 25. x=y2+1

  2. 26. y=x31

In Exercises 2738, graph each equation in a rectangular coordinate system.

  1. 27. y=2x

  2. 28. y=2

  3. 29. x+y=2

  4. 30. y=13 x2

  5. 31. x=3.5

  6. 32. 4x2y=8

  7. 33. f(x)=x24

  8. 34. f(x)=x4

  9. 35. f(x)=|x|4

  10. 36. 5y=3x

  11. 37. 5y=20

  12. 38. f(x)={1ifx02x+1ifx>0

  13. 39. Let f(x)=2x2+x5.

    1. Find f(x). Is f even, odd, or neither?

    2. Find f(x+h)f(x)h , h0.

  14. 40. Let C(x)={30if0x20030+0.40(x200)ifx>200.

    1. Find C(150).

    2. Find C(250).

In Exercises 4144, write a function in slope-intercept form whose graph satisfies the given conditions.

  1. 41. Slope=2, passing through (4, 3)

  2. 42. Passing through (1, 5) and (2, 1)

  3. 43. Passing through (3, 4) and parallel to the line whose equation is 3xy5=0

  4. 44. Passing through (4, 3) and perpendicular to the line whose equation is 2x5y10=0

  5. 45. Determine whether the line through (2, 4) and (7, 0) is parallel to a second line through (4, 2) and (1, 6).

  6. 46. Exercise is useful not only in preventing depression, but also as a treatment. The following graphs show the percentage of patients with depression in remission when exercise (brisk walking) was used as a treatment. (The control group that engaged in no exercise had 11% of the patients in remission.)

    The image shows two graphs titled exercise and the percentage of patients with depression in remission.

    Source: Newsweek, March 26, 2007

    1. Find the slope of the line passing through the two points shown by the voice balloons. Express the slope as a decimal.

    2. Use your answer from part (a) to complete this statement:

      For each minute of brisk walking, the percentage of patients with depression in remission increased by _____%. The rate of change is _____% per _____________.

  7. 47. Find the average rate of change of f(x)=3x2x from x1=1 to x2=2.

Chapter 1: Mid-Chapter Check Point

Chapter 1 Mid-Chapter Check Point

What You Know: We learned that a function is a relation in which no two ordered pairs have the same first component and different second components. We represented functions as equations and used function notation. We graphed functions and applied the vertical line test to identify graphs of functions. We determined the domain and range of a function from its graph, using inputs on the x-axis for the domain and outputs on the y-axis for the range. We used graphs to identify intervals on which functions increase, decrease, or are constant, as well as to locate relative maxima or minima. We determined when graphs of equations are symmetric with respect to the y-axis (no change when x is substituted for x), the x-axis (no change when y is substituted for y), and the origin (no change when x is substituted for x and y is substituted for y). We identified even functions [f(x)=f(x): y-axis symmetry] and odd functions [f(x)=f(x): origin symmetry]. Finally, we studied linear functions and slope, using slope (change in y divided by change in x) to develop various forms for equations of lines:

Point slope form, y minus y sub 1 = m left parenthesis x minus x sub 1 right parenthesis. Slope intercept form, y = f of x = m x + b. Horizontal line, y = f of x = b. Vertical line, x = a. General form, A x + B y + C = 0.

We saw that parallel lines have the same slope and that perpendicular lines have slopes that are negative reciprocals. For linear functions, slope was interpreted as the rate of change of the dependent variable per unit change in the independent variable. For nonlinear functions, the slope of the secant line between (x1, f(x1)) and (x2, f(x2)) described the average rate of change of f from x1 to x2:ΔyΔx=f(x2)f(x1)x2x1.

In Exercises 16, determine whether each relation is a function. Give the domain and range for each relation.

  1. 1. {(2, 6), (1, 4), (2, 6)}

  2. 2. {(0, 1), (2, 1), (3, 4)}

  3. 3.

    A graph is a line that falls from a closed point at (negative 2, 3) through (0, 1) to (1, 0) and rises to an open point at (2, 1).
  4. 4.

    A graph is a continuous curve that starts at a closed point (negative 3, 0) passes through (0, 2), (0, 0), (0, negative 1), and ends at (4, 0).
  5. 5.

    A graph plots six points as follows, (negative 2, negative 2), (negative 1, negative 2), (negative 1, 1), (0, negative 1), (1, 1), and (2, 3).
  6. 6.

    The graph is a curve that falls through (negative 3, 1) and (0, 0) to a closed point at (1, negative 1). All values are estimated.

In Exercises 78, determine whether each equation defines y as a function of x.

  1. 7. x2+y=5

  2. 8. x+y2=5

Use the graph of f to solve Exercises 924. Where applicable, use interval notation.

The graph of y = f of x is a downward opening parabola that rises through (negative 6, 0) with its vertex at (negative 2, 4) and falls through (2, 0). All values are estimated.
  1. 9. Explain why f represents the graph of a function.

  2. 10. Find the domain of f.

  3. 11. Find the range of f.

  4. 12. Find the x-intercept(s).

  5. 13. Find the y-intercept.

  6. 14. Find the interval(s) on which f is increasing.

  7. 15. Find the interval(s) on which f is decreasing.

  8. 16. At what number does f have a relative maximum?

  9. 17. What is the relative maximum of f?

  10. 18. Find f(4).

  11. 19. For what value or values of x is f(x)=2?

  12. 20. For what value or values of x is f(x)=0?

  13. 21. For what values of x is f(x)>0?

  14. 22. Is f(100) positive or negative?

  15. 23. Is f even, odd, or neither?

  16. 24. Find the average rate of change of f from x1=4 to x2=4.

In Exercises 2526, determine whether the graph of each equation is symmetric with respect to the y-axis, the x-axis, the origin, more than one of these, or none of these.

  1. 25. x=y2+1

  2. 26. y=x31

In Exercises 2738, graph each equation in a rectangular coordinate system.

  1. 27. y=2x

  2. 28. y=2

  3. 29. x+y=2

  4. 30. y=13 x2

  5. 31. x=3.5

  6. 32. 4x2y=8

  7. 33. f(x)=x24

  8. 34. f(x)=x4

  9. 35. f(x)=|x|4

  10. 36. 5y=3x

  11. 37. 5y=20

  12. 38. f(x)={1ifx02x+1ifx>0

  13. 39. Let f(x)=2x2+x5.

    1. Find f(x). Is f even, odd, or neither?

    2. Find f(x+h)f(x)h , h0.

  14. 40. Let C(x)={30if0x20030+0.40(x200)ifx>200.

    1. Find C(150).

    2. Find C(250).

In Exercises 4144, write a function in slope-intercept form whose graph satisfies the given conditions.

  1. 41. Slope=2, passing through (4, 3)

  2. 42. Passing through (1, 5) and (2, 1)

  3. 43. Passing through (3, 4) and parallel to the line whose equation is 3xy5=0

  4. 44. Passing through (4, 3) and perpendicular to the line whose equation is 2x5y10=0

  5. 45. Determine whether the line through (2, 4) and (7, 0) is parallel to a second line through (4, 2) and (1, 6).

  6. 46. Exercise is useful not only in preventing depression, but also as a treatment. The following graphs show the percentage of patients with depression in remission when exercise (brisk walking) was used as a treatment. (The control group that engaged in no exercise had 11% of the patients in remission.)

    The image shows two graphs titled exercise and the percentage of patients with depression in remission.

    Source: Newsweek, March 26, 2007

    1. Find the slope of the line passing through the two points shown by the voice balloons. Express the slope as a decimal.

    2. Use your answer from part (a) to complete this statement:

      For each minute of brisk walking, the percentage of patients with depression in remission increased by _____%. The rate of change is _____% per _____________.

  7. 47. Find the average rate of change of f(x)=3x2x from x1=1 to x2=2.

Objective 1: Recognize graphs of common functions

Graphs of Common Functions

  1. Objective 1 Recognize graphs of common functions.

Watch Video

Table 1.4 gives names to seven frequently encountered functions in algebra. The table shows each function’s graph and lists characteristics of the function. Study the shape of each graph and take a few minutes to verify the function’s characteristics from its graph. Knowing these graphs is essential for analyzing their transformations into more complicated graphs.

Table 1.4 Algebra’s Common Graphs

The image shows a table that lists common algebra graphs.
Table 1.4 Full Alternative Text
Objective 2: Use vertical shifts to graph functions

Vertical Shifts

  1. Objective 2 Use vertical shifts to graph functions.

Watch Video

Let’s begin by looking at three graphs whose shapes are the same. Figure 1.60 shows the graphs. The black graph in the middle is the standard quadratic function, f(x)=x2. Now, look at the blue graph on the top. The equation of this graph, g(x)=x2+2, adds 2 to the right side of f(x)=x2. The y-coordinate of each point of g is 2 more than the corresponding y-coordinate of each point of f. What effect does this have on the graph of f? It shifts the graph vertically up by 2 units.

The graph of g is, g of x = x squared + 2 = f of x + 2. 2 is labeled, shifts the graph of f up 2 units.

Figure 1.60 Vertical shifts

The image shows a graph that plots three upward opening parabolas.
Figure 1.60 Full Alternative Text

Finally, look at the red graph on the bottom in Figure 1.60. The equation of this graph, h(x)=x23, subtracts 3 from the right side of f(x)=x2. The y-coordinate of each point of h is 3 less than the corresponding y-coordinate of each point of f. What effect does this have on the graph of f? It shifts the graph vertically down by 3 units.

The graph of h is, h of x = x squared minus 3 = f of x minus 3. 3 is labeled, shifts the graph of f down 3 units.

In general, if c is positive, y=f(x)+c shifts the graph of f upward c units and y=f(x)c shifts the graph of f downward c units. These are called vertical shifts of the graph of f.

Vertical Shifts

Let f be a function and c a positive real number.

The image shows two graphs with two curves.

Example 1 Vertical Shift Downward

Use the graph of f(x)=|x| to obtain the graph of g(x)=|x|4.

Solution

The graph of g(x)=|x|4 has the same shape as the graph of f(x)=|x|. However, it is shifted down vertically 4 units.

The image shows two v shaped graphs.

Check Point 1

  • Use the graph of f(x)=|x| to obtain the graph of g(x)=|x|+3.

Objective 2: Use vertical shifts to graph functions

Vertical Shifts

  1. Objective 2 Use vertical shifts to graph functions.

Watch Video

Let’s begin by looking at three graphs whose shapes are the same. Figure 1.60 shows the graphs. The black graph in the middle is the standard quadratic function, f(x)=x2. Now, look at the blue graph on the top. The equation of this graph, g(x)=x2+2, adds 2 to the right side of f(x)=x2. The y-coordinate of each point of g is 2 more than the corresponding y-coordinate of each point of f. What effect does this have on the graph of f? It shifts the graph vertically up by 2 units.

The graph of g is, g of x = x squared + 2 = f of x + 2. 2 is labeled, shifts the graph of f up 2 units.

Figure 1.60 Vertical shifts

The image shows a graph that plots three upward opening parabolas.
Figure 1.60 Full Alternative Text

Finally, look at the red graph on the bottom in Figure 1.60. The equation of this graph, h(x)=x23, subtracts 3 from the right side of f(x)=x2. The y-coordinate of each point of h is 3 less than the corresponding y-coordinate of each point of f. What effect does this have on the graph of f? It shifts the graph vertically down by 3 units.

The graph of h is, h of x = x squared minus 3 = f of x minus 3. 3 is labeled, shifts the graph of f down 3 units.

In general, if c is positive, y=f(x)+c shifts the graph of f upward c units and y=f(x)c shifts the graph of f downward c units. These are called vertical shifts of the graph of f.

Vertical Shifts

Let f be a function and c a positive real number.

The image shows two graphs with two curves.

Example 1 Vertical Shift Downward

Use the graph of f(x)=|x| to obtain the graph of g(x)=|x|4.

Solution

The graph of g(x)=|x|4 has the same shape as the graph of f(x)=|x|. However, it is shifted down vertically 4 units.

The image shows two v shaped graphs.

Check Point 1

  • Use the graph of f(x)=|x| to obtain the graph of g(x)=|x|+3.

Objective 3: Use horizontal shifts to graph functions

Horizontal Shifts

  1. Objective 3 Use horizontal shifts to graph functions.

Watch Video

We return to the graph of f(x)=x2, the standard quadratic function. In Figure 1.61, the graph of function f is in the middle of the three graphs. By contrast to the vertical shift situation, this time there are graphs to the left and to the right of the graph of f. Look at the blue graph on the right. The equation of this graph, g(x)=(x3)2, subtracts 3 from each value of x before squaring it. What effect does this have on the graph of f(x)=x2? It shifts the graph horizontally to the right by 3 units.

The graph of g is, g of x = left parenthesis x minus 3 right parenthesis squared = f of left parenthesis x minus 3 right parenthesis. f of left parenthesis x minus 3 right parenthesis indicates that it shifts the graph of f, 3 units to the right.

Figure 1.61 Horizontal shifts

A graph plots three upward opening parabolas.
Figure 1.61 Full Alternative Text

Does it seem strange that subtracting 3 in the domain causes a shift of 3 units to the right? Perhaps a partial table of coordinates for each function will numerically convince you of this shift.

x f(x)=x2
2 (2)2=4
1 (1)2=1
0 02=0
1 12=1
2 22=4
x g(x)=(x3)2
1 (13)2=(2)2=4
2 (23)2=(1)2=1
3 (33)2=()02=0
4 (43)2=()12=1
5 (53)2=()22=4

Notice that for the values of f(x) and g(x) to be the same, the values of x used in graphing g must each be 3 units greater than those used to graph f. For this reason, the graph of g is the graph of f shifted 3 units to the right.

Now, look at the red graph on the left in Figure 1.61. The equation of this graph, h(x)=(x+2)2, adds 2 to each value of x before squaring it. What effect does this have on the graph of f(x)=x2? It shifts the graph horizontally to the left by 2 units.

The graph of h is, h of x = left parenthesis x + 2 right parenthesis squared = f of start expression x + 2 end expression. f and 2 indicate that it shifts the graph of f, 2 units to the left.

In general, if c is positive, y=f(x+c) shifts the graph of f to the left c units and y=f(xc) shifts the graph of f to the right c units. These are called horizontal shifts of the graph of f.

Horizontal Shifts

Let f be a function and c a positive real number.

The image shows two graphs.

Example 2 Horizontal Shift to the Left

Use the graph of f(x)=x to obtain the graph of g(x)=x+5.

Solution

Compare the equations for f(x)=x and g(x)=x+5. The equation for g adds 5 to each value of x before taking the square root.

The graph of g is, y = g of x = the square root of start expression x + 5 end expression = f of start expression x + 5 end expression. f of start expression x + 5 end expression indicates that it shifts the graph of f 5 units to the left.

The graph of g(x)=x+5 has the same shape as the graph of f(x)=x. However, it is shifted horizontally to the left 5 units.

The image shows two graphs.

Check Point 2

  • Use the graph of f(x)=x to obtain the graph of g(x)=x4.

Some functions can be graphed by combining horizontal and vertical shifts. These functions will be variations of a function whose equation you know how to graph, such as the standard quadratic function, the standard cubic function, the square root function, the cube root function, or the absolute value function.

In our next example, we will use the graph of the standard quadratic function, f(x)=x2, to obtain the graph of h(x)=(x+1)23. We will graph three functions:

The image shows three functions to be graphed.
1.6-243 Full Alternative Text

Example 3 Combining Horizontal and Vertical Shifts

Use the graph of f(x)=x2 to obtain the graph of h(x)=(x+1)23.

Solution

The image shows three graphs.

Check Point 3

  • Use the graph of f(x)=x to obtain the graph of h(x)=x12.

Objective 3: Use horizontal shifts to graph functions

Horizontal Shifts

  1. Objective 3 Use horizontal shifts to graph functions.

Watch Video

We return to the graph of f(x)=x2, the standard quadratic function. In Figure 1.61, the graph of function f is in the middle of the three graphs. By contrast to the vertical shift situation, this time there are graphs to the left and to the right of the graph of f. Look at the blue graph on the right. The equation of this graph, g(x)=(x3)2, subtracts 3 from each value of x before squaring it. What effect does this have on the graph of f(x)=x2? It shifts the graph horizontally to the right by 3 units.

The graph of g is, g of x = left parenthesis x minus 3 right parenthesis squared = f of left parenthesis x minus 3 right parenthesis. f of left parenthesis x minus 3 right parenthesis indicates that it shifts the graph of f, 3 units to the right.

Figure 1.61 Horizontal shifts

A graph plots three upward opening parabolas.
Figure 1.61 Full Alternative Text

Does it seem strange that subtracting 3 in the domain causes a shift of 3 units to the right? Perhaps a partial table of coordinates for each function will numerically convince you of this shift.

x f(x)=x2
2 (2)2=4
1 (1)2=1
0 02=0
1 12=1
2 22=4
x g(x)=(x3)2
1 (13)2=(2)2=4
2 (23)2=(1)2=1
3 (33)2=()02=0
4 (43)2=()12=1
5 (53)2=()22=4

Notice that for the values of f(x) and g(x) to be the same, the values of x used in graphing g must each be 3 units greater than those used to graph f. For this reason, the graph of g is the graph of f shifted 3 units to the right.

Now, look at the red graph on the left in Figure 1.61. The equation of this graph, h(x)=(x+2)2, adds 2 to each value of x before squaring it. What effect does this have on the graph of f(x)=x2? It shifts the graph horizontally to the left by 2 units.

The graph of h is, h of x = left parenthesis x + 2 right parenthesis squared = f of start expression x + 2 end expression. f and 2 indicate that it shifts the graph of f, 2 units to the left.

In general, if c is positive, y=f(x+c) shifts the graph of f to the left c units and y=f(xc) shifts the graph of f to the right c units. These are called horizontal shifts of the graph of f.

Horizontal Shifts

Let f be a function and c a positive real number.

The image shows two graphs.

Example 2 Horizontal Shift to the Left

Use the graph of f(x)=x to obtain the graph of g(x)=x+5.

Solution

Compare the equations for f(x)=x and g(x)=x+5. The equation for g adds 5 to each value of x before taking the square root.

The graph of g is, y = g of x = the square root of start expression x + 5 end expression = f of start expression x + 5 end expression. f of start expression x + 5 end expression indicates that it shifts the graph of f 5 units to the left.

The graph of g(x)=x+5 has the same shape as the graph of f(x)=x. However, it is shifted horizontally to the left 5 units.

The image shows two graphs.

Check Point 2

  • Use the graph of f(x)=x to obtain the graph of g(x)=x4.

Some functions can be graphed by combining horizontal and vertical shifts. These functions will be variations of a function whose equation you know how to graph, such as the standard quadratic function, the standard cubic function, the square root function, the cube root function, or the absolute value function.

In our next example, we will use the graph of the standard quadratic function, f(x)=x2, to obtain the graph of h(x)=(x+1)23. We will graph three functions:

The image shows three functions to be graphed.
1.6-243 Full Alternative Text

Example 3 Combining Horizontal and Vertical Shifts

Use the graph of f(x)=x2 to obtain the graph of h(x)=(x+1)23.

Solution

The image shows three graphs.

Check Point 3

  • Use the graph of f(x)=x to obtain the graph of h(x)=x12.

Objective 4: Use reflections to graph functions

Reflections of Graphs

  1. Objective 4 Use reflections to graph functions.

Watch Video

This photograph shows a reflection of an old bridge in a river. This perfect reflection occurs because the surface of the water is absolutely still. A mild breeze rippling the water’s surface would distort the reflection.

Is it possible for graphs to have mirror-like qualities? Yes. Figure 1.62 shows the graphs of f(x)=x2 and g(x)=x2. The graph of g is a reflection about the x-axis of the graph of f. For corresponding values of x, the y-coordinates of g are the opposites of the y-coordinates of f. In general, the graph of y=f(x) reflects the graph of f about the x-axis. Thus, the graph of g is a reflection of the graph of f about the x-axis because

g(x)=x2=f(x).

Figure 1.62 Reflection about the x-axis

The image shows a graph plot of two parabolas.
Figure 1.62 Full Alternative Text

Reflection about the x-Axis

The graph of y=f(x) is the graph of y=f(x) reflected about the x-axis.

Example 4 Reflection about the x-Axis

Use the graph of f(x)=x3 to obtain the graph of g(x)=x3.

Solution

Compare the equations for f(x)=x3 and g(x)=x3. The graph of g is a reflection about the x-axis of the graph of f because

g(x)=x3=f(x).
The image shows two graphs.

Check Point 4

  • Use the graph of f(x)=|x| to obtain the graph of g(x)=|x|.

It is also possible to reflect graphs about the y-axis.

Reflection about the y-Axis

The graph of y=f(x) is the graph of y=f(x) reflected about the y-axis.

For each point (x, y) on the graph of y=f(x), the point (x, y) is on the graph of y=f(x).

Example 5 Reflection about the y-Axis

Use the graph of f(x)=x to obtain the graph of h(x)=x.

Solution

Compare the equations for f(x)=x and h(x)=x. The graph of h is a reflection about the y-axis of the graph of f because

h(x)=x=f(x).
The image shows two graphs.

Check Point 5

  • Use the graph of f(x)=x3 to obtain the graph of h(x)=x3.

Objective 5: Use vertical stretching and shrinking to graph functions

Vertical Stretching and Shrinking

  1. Objective 5 Use vertical stretching and shrinking to graph functions.

Watch Video

Morphing does much more than move an image horizontally, vertically, or about an axis. An object having one shape is transformed into a different shape. Horizontal shifts, vertical shifts, and reflections do not change the basic shape of a graph. Graphs remain rigid and proportionally the same when they undergo these transformations. How can we shrink and stretch graphs, thereby altering their basic shapes?

Look at the three graphs in Figure 1.63. The black graph in the middle is the graph of the standard quadratic function, f(x)=x2. Now, look at the blue graph on the top. The equation of this graph is g(x)=2x2, or g(x)=2f(x). Thus, for each x, the y-coordinate of g is two times as large as the corresponding y-coordinate on the graph of f. The result is a narrower graph because the values of y are rising faster. We say that the graph of g is obtained by vertically stretching the graph of f. Now, look at the red graph on the bottom. The equation of this graph is h(x)=12 x2, or h(x)=12 f(x). Thus, for each x, the y-coordinate of h is one-half as large as the corresponding y-coordinate on the graph of f. The result is a wider graph because the values of y are rising more slowly. We say that the graph of h is obtained by vertically shrinking the graph of f.

Figure 1.63 Vertically stretching and shrinking f(x)=x2

A graph plots three upward opening parabolas.
Figure 1.63 Full Alternative Text

These observations can be summarized as follows:

Vertically Stretching and Shrinking Graphs

Let f be a function and c a positive real number.

The image shows two graphs.

Example 6 Vertically Shrinking a Graph

Use the graph of f(x)=x3 to obtain the graph of h(x)=12 x3.

Solution

The graph of h(x)=12 x3 is obtained by vertically shrinking the graph of f(x)=x3.

The image shows two graphs.

Check Point 6

  • Use the graph of f(x)=|x| to obtain the graph of g(x)=2|x|.

Objective 5: Use vertical stretching and shrinking to graph functions

Vertical Stretching and Shrinking

  1. Objective 5 Use vertical stretching and shrinking to graph functions.

Watch Video

Morphing does much more than move an image horizontally, vertically, or about an axis. An object having one shape is transformed into a different shape. Horizontal shifts, vertical shifts, and reflections do not change the basic shape of a graph. Graphs remain rigid and proportionally the same when they undergo these transformations. How can we shrink and stretch graphs, thereby altering their basic shapes?

Look at the three graphs in Figure 1.63. The black graph in the middle is the graph of the standard quadratic function, f(x)=x2. Now, look at the blue graph on the top. The equation of this graph is g(x)=2x2, or g(x)=2f(x). Thus, for each x, the y-coordinate of g is two times as large as the corresponding y-coordinate on the graph of f. The result is a narrower graph because the values of y are rising faster. We say that the graph of g is obtained by vertically stretching the graph of f. Now, look at the red graph on the bottom. The equation of this graph is h(x)=12 x2, or h(x)=12 f(x). Thus, for each x, the y-coordinate of h is one-half as large as the corresponding y-coordinate on the graph of f. The result is a wider graph because the values of y are rising more slowly. We say that the graph of h is obtained by vertically shrinking the graph of f.

Figure 1.63 Vertically stretching and shrinking f(x)=x2

A graph plots three upward opening parabolas.
Figure 1.63 Full Alternative Text

These observations can be summarized as follows:

Vertically Stretching and Shrinking Graphs

Let f be a function and c a positive real number.

The image shows two graphs.

Example 6 Vertically Shrinking a Graph

Use the graph of f(x)=x3 to obtain the graph of h(x)=12 x3.

Solution

The graph of h(x)=12 x3 is obtained by vertically shrinking the graph of f(x)=x3.

The image shows two graphs.

Check Point 6

  • Use the graph of f(x)=|x| to obtain the graph of g(x)=2|x|.

Objective 6: Use horizontal stretching and shrinking to graph functions

Horizontal Stretching and Shrinking

  1. Objective 6 Use horizontal stretching and shrinking to graph functions.

Watch Video

It is also possible to stretch and shrink graphs horizontally.

Horizontally Stretching and Shrinking Graphs

Let f be a function and c a positive real number.

The image shows two graphs.

Example 7 Horizontally Stretching and Shrinking a Graph

Use the graph of y=f(x) in Figure 1.64 to obtain each of the following graphs:

  1. g(x)=f(2x)

  2. h(x)=f(12 x).

Figure 1.64

The graph of y = f of x is a curve that starts at (negative 4, 0), passes through (negative 2, 4), (0, 0), (2, negative 2), and ends at (4, 0).

Solution

  1. The graph of g(x)=f(2x) is obtained by horizontally shrinking the graph of y=f(x).

    The image shows two graphs.

  2. The graph of h(x)=f(12 x) is obtained by horizontally stretching the graph of y=f(x).

The image shows two graphs

Check Point 7

  • Use the graph of y=f(x) in Figure 1.65 to obtain each of the following graphs:

    1. g(x)=f(2x)

    2. h(x)=f(12 x).

    Figure 1.65

    The graph of y = f of x is a curve that starts at (negative 2, 0), passes through (0, 3), (2, 0), (4, negative 3), and ends at (6, 0).
Objective 7: Graph functions involving a sequence of transformations

Sequences of Transformations

  1. Objective 7 Graph functions involving a sequence of transformations.

Watch Video

Table 1.5 summarizes the procedures for transforming the graph of y=f(x).

Table 1.5 Summary of Transformations

To Graph: Draw the Graph of f and: Changes in the Equation of y=f(x)

Vertical shifts

y=f(x)+c

y=f(x)c

 

Raise the graph of f by c units.

Lower the graph of f by c units.

 

c is added to f(x).

c is subtracted from f(x)

Horizontal shifts

y=f(x+c)

y=f(xc)

 

Shift the graph of f to the left c units.

Shift the graph of f to the right c units.

 

x is replaced with x+c.

x is replaced with xc.

Reflection about the x-axis

y=f(x)

Reflect the graph of f about the x-axis. f(x) is multiplied by 1.

Reflection about the y-axis

y=f(x)

Reflect the graph of f about the y-axis. x is replaced with x.

Vertical stretching or shrinking

y=cf (x), c>1

y=cf (x), 0<c<1

 

Multiply each y-coordinate of y=f(x) by c, vertically stretching the graph of f.

Multiply each y-coordinate of y=f(x) by c, vertically shrinking the graph of f.

 

f(x) is multiplied by c, c>1.

f(x) is multiplied by c, 0<c<1.

Horizontal stretching or shrinking

y=f(cx), c>1

y=f(cx), 0<c<1

 

Divide each x-coordinate of y=f(x) by c, horizontally shrinking the graph of f.

Divide each x-coordinate of y=f(x) by c, horizontally stretching the graph of f.

 

x is replaced with cx, c>1.

x is replaced with cx, 0<c<1.

In each case, c represents a positive real number.

Order of Transformations

A function involving more than one transformation can be graphed by performing transformations in the following order:

  1. Horizontal shifting

  2. Stretching or shrinking

  3. Reflecting

  4. Vertical shifting

Example 8 Graphing Using a Sequence of Transformations

Use the graph of y=f(x) given in Figure 1.64 of Example 7, and repeated below, to graph y= 12 f(x1)+3.

Solution

Our graphs will evolve in the following order:

  1. Horizontal shifting: Graph y=f(x1) by shifting the graph of y=f(x) 1 unit to the right.

  2. Shrinking: Graph y=12 f(x1) by shrinking the previous graph by a factor of 12.

  3. Reflecting: Graph y= 12 f(x1) by reflecting the previous graph about the x-axis.

  4. Vertical shifting: Graph y= 12 f(x1)+3 by shifting the previous graph up 3 units.

The image shows five graphs.

Check Point 8

Example 9 Graphing Using a Sequence of Transformations

Use the graph of f(x)=x2 to graph g(x)=2(x+3)21.

Solution

Our graphs will evolve in the following order:

  1. Horizontal shifting: Graph y=(x+3)2 by shifting the graph of f(x)=x2 three units to the left.

  2. Stretching: Graph y=2(x+3)2 by stretching the previous graph by a factor of 2.

  3. Vertical shifting: Graph g(x)=2(x+3)21 by shifting the previous graph down 1 unit.

The image shows four graphs.

Check Point 9

  • Use the graph of f(x)=x2 to graph g(x)=2(x1)2+3.

Objective 7: Graph functions involving a sequence of transformations

Sequences of Transformations

  1. Objective 7 Graph functions involving a sequence of transformations.

Watch Video

Table 1.5 summarizes the procedures for transforming the graph of y=f(x).

Table 1.5 Summary of Transformations

To Graph: Draw the Graph of f and: Changes in the Equation of y=f(x)

Vertical shifts

y=f(x)+c

y=f(x)c

 

Raise the graph of f by c units.

Lower the graph of f by c units.

 

c is added to f(x).

c is subtracted from f(x)

Horizontal shifts

y=f(x+c)

y=f(xc)

 

Shift the graph of f to the left c units.

Shift the graph of f to the right c units.

 

x is replaced with x+c.

x is replaced with xc.

Reflection about the x-axis

y=f(x)

Reflect the graph of f about the x-axis. f(x) is multiplied by 1.

Reflection about the y-axis

y=f(x)

Reflect the graph of f about the y-axis. x is replaced with x.

Vertical stretching or shrinking

y=cf (x), c>1

y=cf (x), 0<c<1

 

Multiply each y-coordinate of y=f(x) by c, vertically stretching the graph of f.

Multiply each y-coordinate of y=f(x) by c, vertically shrinking the graph of f.

 

f(x) is multiplied by c, c>1.

f(x) is multiplied by c, 0<c<1.

Horizontal stretching or shrinking

y=f(cx), c>1

y=f(cx), 0<c<1

 

Divide each x-coordinate of y=f(x) by c, horizontally shrinking the graph of f.

Divide each x-coordinate of y=f(x) by c, horizontally stretching the graph of f.

 

x is replaced with cx, c>1.

x is replaced with cx, 0<c<1.

In each case, c represents a positive real number.

Order of Transformations

A function involving more than one transformation can be graphed by performing transformations in the following order:

  1. Horizontal shifting

  2. Stretching or shrinking

  3. Reflecting

  4. Vertical shifting

Example 8 Graphing Using a Sequence of Transformations

Use the graph of y=f(x) given in Figure 1.64 of Example 7, and repeated below, to graph y= 12 f(x1)+3.

Solution

Our graphs will evolve in the following order:

  1. Horizontal shifting: Graph y=f(x1) by shifting the graph of y=f(x) 1 unit to the right.

  2. Shrinking: Graph y=12 f(x1) by shrinking the previous graph by a factor of 12.

  3. Reflecting: Graph y= 12 f(x1) by reflecting the previous graph about the x-axis.

  4. Vertical shifting: Graph y= 12 f(x1)+3 by shifting the previous graph up 3 units.

The image shows five graphs.

Check Point 8

Example 9 Graphing Using a Sequence of Transformations

Use the graph of f(x)=x2 to graph g(x)=2(x+3)21.

Solution

Our graphs will evolve in the following order:

  1. Horizontal shifting: Graph y=(x+3)2 by shifting the graph of f(x)=x2 three units to the left.

  2. Stretching: Graph y=2(x+3)2 by stretching the previous graph by a factor of 2.

  3. Vertical shifting: Graph g(x)=2(x+3)21 by shifting the previous graph down 1 unit.

The image shows four graphs.

Check Point 9

  • Use the graph of f(x)=x2 to graph g(x)=2(x1)2+3.

Objective 7: Graph functions involving a sequence of transformations

Sequences of Transformations

  1. Objective 7 Graph functions involving a sequence of transformations.

Watch Video

Table 1.5 summarizes the procedures for transforming the graph of y=f(x).

Table 1.5 Summary of Transformations

To Graph: Draw the Graph of f and: Changes in the Equation of y=f(x)

Vertical shifts

y=f(x)+c

y=f(x)c

 

Raise the graph of f by c units.

Lower the graph of f by c units.

 

c is added to f(x).

c is subtracted from f(x)

Horizontal shifts

y=f(x+c)

y=f(xc)

 

Shift the graph of f to the left c units.

Shift the graph of f to the right c units.

 

x is replaced with x+c.

x is replaced with xc.

Reflection about the x-axis

y=f(x)

Reflect the graph of f about the x-axis. f(x) is multiplied by 1.

Reflection about the y-axis

y=f(x)

Reflect the graph of f about the y-axis. x is replaced with x.

Vertical stretching or shrinking

y=cf (x), c>1

y=cf (x), 0<c<1

 

Multiply each y-coordinate of y=f(x) by c, vertically stretching the graph of f.

Multiply each y-coordinate of y=f(x) by c, vertically shrinking the graph of f.

 

f(x) is multiplied by c, c>1.

f(x) is multiplied by c, 0<c<1.

Horizontal stretching or shrinking

y=f(cx), c>1

y=f(cx), 0<c<1

 

Divide each x-coordinate of y=f(x) by c, horizontally shrinking the graph of f.

Divide each x-coordinate of y=f(x) by c, horizontally stretching the graph of f.

 

x is replaced with cx, c>1.

x is replaced with cx, 0<c<1.

In each case, c represents a positive real number.

Order of Transformations

A function involving more than one transformation can be graphed by performing transformations in the following order:

  1. Horizontal shifting

  2. Stretching or shrinking

  3. Reflecting

  4. Vertical shifting

Example 8 Graphing Using a Sequence of Transformations

Use the graph of y=f(x) given in Figure 1.64 of Example 7, and repeated below, to graph y= 12 f(x1)+3.

Solution

Our graphs will evolve in the following order:

  1. Horizontal shifting: Graph y=f(x1) by shifting the graph of y=f(x) 1 unit to the right.

  2. Shrinking: Graph y=12 f(x1) by shrinking the previous graph by a factor of 12.

  3. Reflecting: Graph y= 12 f(x1) by reflecting the previous graph about the x-axis.

  4. Vertical shifting: Graph y= 12 f(x1)+3 by shifting the previous graph up 3 units.

The image shows five graphs.

Check Point 8

Example 9 Graphing Using a Sequence of Transformations

Use the graph of f(x)=x2 to graph g(x)=2(x+3)21.

Solution

Our graphs will evolve in the following order:

  1. Horizontal shifting: Graph y=(x+3)2 by shifting the graph of f(x)=x2 three units to the left.

  2. Stretching: Graph y=2(x+3)2 by stretching the previous graph by a factor of 2.

  3. Vertical shifting: Graph g(x)=2(x+3)21 by shifting the previous graph down 1 unit.

The image shows four graphs.

Check Point 9

  • Use the graph of f(x)=x2 to graph g(x)=2(x1)2+3.

1.6: Exercise Set

1.6 Exercise Set

Practice Exercises

In Exercises 116, use the graph of y=f(x) to graph each function g.

The graph of y = f of x is a line segment that starts at (negative 2, 2), passes through (0, 2), and ends at (2, 2).
  1. 1. g(x)=f(x)+1

  2. 2. g(x)=f(x)1

  3. 3. g(x)=f(x+1)

  4. 4. g(x)=f(x1)

  5. 5. g(x)=f(x1)2

  6. 6. g(x)=f(x+1)+2

  7. 7. g(x)=f(x)

  8. 8. g(x)=f(x)

  9. 9. g(x)=f(x)+3

  10. 10. g(x)=f(x)+3

  11. 11. g(x)=12 f(x)

  12. 12. g(x)=2f(x)

  13. 13. g(x)=f(12 x)

  14. 14. g(x)=f(2x)

  15. 15. g(x)=f(12 x)+1

  16. 16. g(x)=f(2x)1

In Exercises 1732, use the graph of y=f(x) to graph each function g.

The graph of y = f of x is a curve that starts at (negative 4, 0), passes through (negative 2, negative 2), (0, 0), (2, 2), and ends at (4, 0).
  1. 17. g(x)=f(x)1

  2. 18. g(x)=f(x)+1

  3. 19. g(x)=f(x1)

  4. 20. g(x)=f(x+1)

  5. 21. g(x)=f(x1)+2

  6. 22. g(x)=f(x+1)2

  7. 23. g(x)=f(x)

  8. 24. g(x)=f(x)

  9. 25. g(x)=f(x)+1

  10. 26. g(x)=f(x)+1

  11. 27. g(x)=2f(x)

  12. 28. g(x)=12 f(x)

  13. 29. g(x)=f(2x)

  14. 30. g(x)=f(12 x)

  15. 31. g(x)=2f(x+2)+1

  16. 32. g(x)=2f(x+2)1

In Exercises 3344, use the graph of y=f(x) to graph each function g.

A graph plots two lines for y = f of x. The first line falls from a closed point (0, 0) to an open point (4, negative 2). The second line falls from a closed point (negative 4, 0) to an open point (0, negative 2).
  1. 33. g(x)=f(x)+2

  2. 34. g(x)=f(x)2

  3. 35. g(x)=f(x+2)

  4. 36. g(x)=f(x2)

  5. 37. g(x)=f(x+2)

  6. 38. g(x)=f(x2)

  7. 39. g(x)= 12f(x+2)

  8. 40. g(x)= 12f(x2)

  9. 41. g(x)= 12f(x+2)2

  10. 42. g(x)= 12f(x2)+2

  11. 43. g(x)=12f(2x)

  12. 44. g(x)=2f(12 x)

In Exercises 4552, use the graph of y=f(x) to graph each function g.

A graph of y = f of x is a continuous line that starts from (negative 4, negative 2) passes through (negative 2, 0), (0, 2) (2, 2) and ends at (4, 0).
  1. 45. g(x)=f(x1)1

  2. 46. g(x)=f(x+1)+1

  3. 47. g(x)=f(x1)+1

  4. 48. g(x)=f(x+1)1

  5. 49. g(x)=2f(12 x)

  6. 50. g(x)=12f(2x)

  7. 51. g(x)=12f(x+1)

  8. 52. g(x)=2f(x1)

In Exercises 5366, begin by graphing the standard quadratic function, f(x)=x2. Then use transformations of this graph to graph the given function.

  1. 53. g(x)=x22

  2. 54. g(x)=x21

  3. 55. g(x)=(x2)2

  4. 56. g(x)=(x1)2

  5. 57. h(x)=(x2)2

  6. 58. h(x)=(x1)2

  7. 59. h(x)=(x2)2+1

  8. 60. h(x)=(x1)2+2

  9. 61. g(x)=2(x2)2

  10. 62. g(x)=12 (x1)2

  11. 63. h(x)=2(x2)21

  12. 64. h(x)=12 (x1)21

  13. 65. h(x)=2(x+1)2+1

  14. 66. h(x)=2(x+2)2+1

In Exercises 6780, begin by graphing the square root function, f(x)=x. Then use transformations of this graph to graph the given function.

  1. 67. g(x)=x+2

  2. 68. g(x)=x+1

  3. 69. g(x)=x+2

  4. 70. g(x)=x+1

  5. 71. h(x)=x+2

  6. 72. h(x)=x+1

  7. 73. h(x)=x+2

  8. 74. h(x)=x+1

  9. 75. g(x)=12x+2

  10. 76. g(x)=2x+1

  11. 77. h(x)=x+22

  12. 78. h(x)=x+11

  13. 79. g(x)=2x+22

  14. 80. g(x)=2x+11

In Exercises 8194, begin by graphing the absolute value function, f(x)=|x|. Then use transformations of this graph to graph the given function.

  1. 81. g(x)=|x|+4

  2. 82. g(x)=|x|+3

  3. 83. g(x)=|x+4|

  4. 84. g(x)=|x+3|

  5. 85. h(x)=|x+4|2

  6. 86. h(x)=|x+3|2

  7. 87. h(x)=|x+4|

  8. 88. h(x)=|x+3|

  9. 89. g(x)=|x+4|+1

  10. 90. g(x)=|x+4|+2

  11. 91. h(x)=2|x+4|

  12. 92. h(x)=2|x+3|

  13. 93. g(x)=2|x+4|+1

  14. 94. g(x)=2|x+3|+2

In Exercises 95106, begin by graphing the standard cubic function, f(x)=x3. Then use transformations of this graph to graph the given function.

  1. 95. g(x)=x33

  2. 96. g(x)=x32

  3. 97. g(x)=(x3)3

  4. 98. g(x)=(x2)3

  5. 99. h(x)=x3

  6. 100. h(x)=(x2)3

  7. 101. h(x)=12 x3

  8. 102. h(x)=14 x3

  9. 103. r(x)=(x3)3+2

  10. 104. r(x)=(x2)3+1

  11. 105. h(x)=12 (x3)32

  12. 106. h(x)=12 (x2)31

In Exercises 107118, begin by graphing the cube root function, f(x)=x3. Then use transformations of this graph to graph the given function.

  1. 107. g(x)=x3+2

  2. 108. g(x)=x32

  3. 109. g(x)=x+23

  4. 110. g(x)=x23

  5. 111. h(x)=12x+23

  6. 112. h(x)=12x23

  7. 113. r(x)=12x+232

  8. 114. r(x)=12x23+2

  9. 115. h(x)=x+23

  10. 116. h(x)=x23

  11. 117. g(x)=x23

  12. 118. g(x)=x+23

Practice PLUS

In Exercises 119122, use transformations of the graph of the greatest integer function, f(x)=int(x) to graph each function. (The graph of f(x)=int(x) is shown in Figure 1.39.)

  1. 119. g(x)=2 int (x+1)

  2. 120. g(x)=3 int (x1)

  3. 121. h(x)=int(x)+1

  4. 122. h(x)=int(x)1

In Exercises 123126, write a possible equation for the function whose graph is shown. Each graph shows a transformation of a common function.

  1. 123.

    A viewing window displays a concave down increasing curve that rises from (2, 0), passes through (5.5, 2), and (8, 2.5). All values are estimated.
  2. 124.

    A viewing window displays a curve that passes through (negative 6, 6), (negative 3, 3), (0, 2), (3.5, 0), and (6, negative 6). All values are estimated.
  3. 125.

    A viewing window displays an upward opening parabola that falls through (negative 8, 5), (negative 6, 0) with its vertex at (negative 3, negative 4) and rises through (2, 0), (5, 8). All values are estimated.
  4. 126.

    A viewing window displays a concave down increasing curve that rises from (2, 1), passes through (6, 3), and (9, 3.5). All values are estimated.

Application Exercises

  1. 127. The function f(x)=2.9x+20.1 models the median height, f(x), in inches, of boys who are x months of age. The graph of f is shown.

    A graph titled, boys height plots age in months versus median height in inches. The horizontal axis ranges from 0 to 60 in increments of 10, and the vertical axis ranges from 0 to 50 in increments of 10.

    Source: Laura Walther Nathanson, The Portable Pediatrician for Parents

    1. Describe how the graph can be obtained using transformations of the square root function f(x)=x.

    2. According to the model, what is the median height of boys who are 48 months, or four years, old? Use a calculator and round to the nearest tenth of an inch. The actual median height for boys at 48 months is 40.8 inches. How well does the model describe the actual height?

    3. Use the model to find the average rate of change, in inches per month, between birth and 10 months. Round to the nearest tenth.

    4. Use the model to find the average rate of change, in inches per month, between 50 and 60 months. Round to the nearest tenth. How does this compare with your answer in part (c)? How is this difference shown by the graph?

  2. 128. The function f(x)=3.1x+19 models the median height, f(x), in inches, of girls who are x months of age. The graph of f is shown.

    A graph titled, girls height plots age in months versus median height in inches. The horizontal axis ranges from 0 to 60 in increments of 10, and the vertical axis ranges from 0 to 50 in increments of 10.

    Source: Laura Walther Nathanson, The Portable Pediatrician for Parents

    1. Describe how the graph can be obtained using transformations of the square root function f(x)=x.

    2. According to the model, what is the median height of girls who are 48 months, or 4 years, old? Use a calculator and round to the nearest tenth of an inch. The actual median height for girls at 48 months is 40.2 inches. How well does the model describe the actual height?

    3. Use the model to find the average rate of change, in inches per month, between birth and 10 months. Round to the nearest tenth.

    4. Use the model to find the average rate of change, in inches per month, between 50 and 60 months. Round to the nearest tenth. How does this compare with your answer in part (c)? How is this difference shown by the graph?

Explaining the Concepts

  1. 129. What must be done to a function’s equation so that its graph is shifted vertically upward?

  2. 130. What must be done to a function’s equation so that its graph is shifted horizontally to the right?

  3. 131. What must be done to a function’s equation so that its graph is reflected about the x-axis?

  4. 132. What must be done to a function’s equation so that its graph is reflected about the y-axis?

  5. 133. What must be done to a function’s equation so that its graph is stretched vertically?

  6. 134. What must be done to a function’s equation so that its graph is shrunk horizontally?

Technology Exercises

  1. 135.

    1. Use a graphing utility to graph f(x)=x2+1.

    2. Graph f(x)=x2+1, g(x)=f(2x), h(x)=f(3x), and k(x)=f(4x) in the same viewing rectangle.

    3. Describe the relationship among the graphs of f, g, h, and k, with emphasis on different values of x for points on all four graphs that give the same y-coordinate.

    4. Generalize by describing the relationship between the graph of f and the graph of g, where g(x)=f(cx) for c>1.

    5. Try out your generalization by sketching the graphs of f(cx) for c=1, c=2, c=3, and c=4 for a function of your choice.

  2. 136.

    1. Use a graphing utility to graph f(x)=x2+1.

    2. Graph f(x)=x2+1, g(x)=f(12 x), and h(x)=f(14 x) in the same viewing rectangle.

    3. Describe the relationship among the graphs of f, g, and h, with emphasis on different values of x for points on all three graphs that give the same y-coordinate.

    4. Generalize by describing the relationship between the graph of f and the graph of g, where g(x)=f(cx) for 0<c<1.

    5. Try out your generalization by sketching the graphs of f(cx) for c=1, and c=12, and c=14 for a function of your choice.

Critical Thinking Exercises

Make Sense? During the winter, you program your home thermostat so that at midnight, the temperature is 55°. This temperature is maintained until 6 a.m. Then the house begins to warm up so that by 9 a.m. the temperature is 65°. At 6 p.m. the house begins to cool. By 9 p.m., the temperature is again 55°. The graph illustrates home temperature, f(t), as a function of hours after midnight, t.

A graph titled, home temperature as a function of time plots hours after midnight versus temperature in degree Fahrenheit.

In Exercises 137140, determine whether each statement makes sense or does not make sense, and explain your reasoning. If the statement makes sense, graph the new function on the domain [0, 24]. If the statement does not make sense, correct the function in the statement and graph the corrected function on the domain [0, 24].

  1. 137. I decided to keep the house 5° warmer than before, so I reprogrammed the thermostat to y=f(t)+5.

  2. 138. I decided to keep the house 5° cooler than before, so I reprogrammed the thermostat to y=f(t)5.

  3. 139. I decided to change the heating schedule to start one hour earlier than before, so I reprogrammed the thermostat to y=f(t1).

  4. 140. I decided to change the heating schedule to start one hour later than before, so I reprogrammed the thermostat to y=f(t+1).

In Exercises 141144, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 141. If f(x)=|x| and g(x)=|x+3|+3, then the graph of g is a translation of the graph of f 3 units to the right and 3 units upward.

  2. 142. If f(x)=x and g(x)=x, then f and g have identical graphs.

  3. 143. If f(x)=x2 and g(x)=5(x22), then the graph of g can be obtained from the graph of f by stretching f 5 units followed by a downward shift of 2 units.

  4. 144. If f(x)=x3 and g(x)=(x3)34, then the graph of g can be obtained from the graph of f by moving f 3 units to the right, reflecting about the x-axis, and then moving the resulting graph down 4 units.

In Exercises 145148, functions f and g are graphed in the same rectangular coordinate system. If g is obtained from f through a sequence of transformations, find an equation for g.

  1. 145.

    The image shows the graph plots two parabolas.
  2. 146.

    The image shows a graph that plots two v shaped curves.
  3. 147.

    A graph plots two curves.
  4. 148.

    The image shows a graph that plots a semicircle and a curve.

For Exercises 149152, assume that (a, b) is a point on the graph of f. What is the corresponding point on the graph of each of the following functions?

  1. 149. y=f(x)

  2. 150. y=2f(x)

  3. 151. y=f(x3)

  4. 152. y=f(x)3

Retaining the Concepts

  1. 153. The length of a rectangle exceeds the width by 13 yards. If the perimeter of the rectangle is 82 yards, what are its dimensions?

    (Section P.8, Example 6)

  2. 154. Solve: x+104=x.

    (Section P.7, Example 12)

  3. 155. if f(x)=x2+3x+2, find f(x+h)f(x)h, h0, and simplify.

    (Section 1.3, Example 8)

Preview Exercises

Exercises 156158 will help you prepare for the material covered in the next section.

In Exercises 156157, perform the indicated operation or operations.

  1. 156. (2x1)(x2+x2)

  2. 157. (f(x))22f(x)+6, where f(x)=3x4

  3. 158. Simplify: 23x1.

1.6: Exercise Set

1.6 Exercise Set

Practice Exercises

In Exercises 116, use the graph of y=f(x) to graph each function g.

The graph of y = f of x is a line segment that starts at (negative 2, 2), passes through (0, 2), and ends at (2, 2).
  1. 1. g(x)=f(x)+1

  2. 2. g(x)=f(x)1

  3. 3. g(x)=f(x+1)

  4. 4. g(x)=f(x1)

  5. 5. g(x)=f(x1)2

  6. 6. g(x)=f(x+1)+2

  7. 7. g(x)=f(x)

  8. 8. g(x)=f(x)

  9. 9. g(x)=f(x)+3

  10. 10. g(x)=f(x)+3

  11. 11. g(x)=12 f(x)

  12. 12. g(x)=2f(x)

  13. 13. g(x)=f(12 x)

  14. 14. g(x)=f(2x)

  15. 15. g(x)=f(12 x)+1

  16. 16. g(x)=f(2x)1

In Exercises 1732, use the graph of y=f(x) to graph each function g.

The graph of y = f of x is a curve that starts at (negative 4, 0), passes through (negative 2, negative 2), (0, 0), (2, 2), and ends at (4, 0).
  1. 17. g(x)=f(x)1

  2. 18. g(x)=f(x)+1

  3. 19. g(x)=f(x1)

  4. 20. g(x)=f(x+1)

  5. 21. g(x)=f(x1)+2

  6. 22. g(x)=f(x+1)2

  7. 23. g(x)=f(x)

  8. 24. g(x)=f(x)

  9. 25. g(x)=f(x)+1

  10. 26. g(x)=f(x)+1

  11. 27. g(x)=2f(x)

  12. 28. g(x)=12 f(x)

  13. 29. g(x)=f(2x)

  14. 30. g(x)=f(12 x)

  15. 31. g(x)=2f(x+2)+1

  16. 32. g(x)=2f(x+2)1

In Exercises 3344, use the graph of y=f(x) to graph each function g.

A graph plots two lines for y = f of x. The first line falls from a closed point (0, 0) to an open point (4, negative 2). The second line falls from a closed point (negative 4, 0) to an open point (0, negative 2).
  1. 33. g(x)=f(x)+2

  2. 34. g(x)=f(x)2

  3. 35. g(x)=f(x+2)

  4. 36. g(x)=f(x2)

  5. 37. g(x)=f(x+2)

  6. 38. g(x)=f(x2)

  7. 39. g(x)= 12f(x+2)

  8. 40. g(x)= 12f(x2)

  9. 41. g(x)= 12f(x+2)2

  10. 42. g(x)= 12f(x2)+2

  11. 43. g(x)=12f(2x)

  12. 44. g(x)=2f(12 x)

In Exercises 4552, use the graph of y=f(x) to graph each function g.

A graph of y = f of x is a continuous line that starts from (negative 4, negative 2) passes through (negative 2, 0), (0, 2) (2, 2) and ends at (4, 0).
  1. 45. g(x)=f(x1)1

  2. 46. g(x)=f(x+1)+1

  3. 47. g(x)=f(x1)+1

  4. 48. g(x)=f(x+1)1

  5. 49. g(x)=2f(12 x)

  6. 50. g(x)=12f(2x)

  7. 51. g(x)=12f(x+1)

  8. 52. g(x)=2f(x1)

In Exercises 5366, begin by graphing the standard quadratic function, f(x)=x2. Then use transformations of this graph to graph the given function.

  1. 53. g(x)=x22

  2. 54. g(x)=x21

  3. 55. g(x)=(x2)2

  4. 56. g(x)=(x1)2

  5. 57. h(x)=(x2)2

  6. 58. h(x)=(x1)2

  7. 59. h(x)=(x2)2+1

  8. 60. h(x)=(x1)2+2

  9. 61. g(x)=2(x2)2

  10. 62. g(x)=12 (x1)2

  11. 63. h(x)=2(x2)21

  12. 64. h(x)=12 (x1)21

  13. 65. h(x)=2(x+1)2+1

  14. 66. h(x)=2(x+2)2+1

In Exercises 6780, begin by graphing the square root function, f(x)=x. Then use transformations of this graph to graph the given function.

  1. 67. g(x)=x+2

  2. 68. g(x)=x+1

  3. 69. g(x)=x+2

  4. 70. g(x)=x+1

  5. 71. h(x)=x+2

  6. 72. h(x)=x+1

  7. 73. h(x)=x+2

  8. 74. h(x)=x+1

  9. 75. g(x)=12x+2

  10. 76. g(x)=2x+1

  11. 77. h(x)=x+22

  12. 78. h(x)=x+11

  13. 79. g(x)=2x+22

  14. 80. g(x)=2x+11

In Exercises 8194, begin by graphing the absolute value function, f(x)=|x|. Then use transformations of this graph to graph the given function.

  1. 81. g(x)=|x|+4

  2. 82. g(x)=|x|+3

  3. 83. g(x)=|x+4|

  4. 84. g(x)=|x+3|

  5. 85. h(x)=|x+4|2

  6. 86. h(x)=|x+3|2

  7. 87. h(x)=|x+4|

  8. 88. h(x)=|x+3|

  9. 89. g(x)=|x+4|+1

  10. 90. g(x)=|x+4|+2

  11. 91. h(x)=2|x+4|

  12. 92. h(x)=2|x+3|

  13. 93. g(x)=2|x+4|+1

  14. 94. g(x)=2|x+3|+2

In Exercises 95106, begin by graphing the standard cubic function, f(x)=x3. Then use transformations of this graph to graph the given function.

  1. 95. g(x)=x33

  2. 96. g(x)=x32

  3. 97. g(x)=(x3)3

  4. 98. g(x)=(x2)3

  5. 99. h(x)=x3

  6. 100. h(x)=(x2)3

  7. 101. h(x)=12 x3

  8. 102. h(x)=14 x3

  9. 103. r(x)=(x3)3+2

  10. 104. r(x)=(x2)3+1

  11. 105. h(x)=12 (x3)32

  12. 106. h(x)=12 (x2)31

In Exercises 107118, begin by graphing the cube root function, f(x)=x3. Then use transformations of this graph to graph the given function.

  1. 107. g(x)=x3+2

  2. 108. g(x)=x32

  3. 109. g(x)=x+23

  4. 110. g(x)=x23

  5. 111. h(x)=12x+23

  6. 112. h(x)=12x23

  7. 113. r(x)=12x+232

  8. 114. r(x)=12x23+2

  9. 115. h(x)=x+23

  10. 116. h(x)=x23

  11. 117. g(x)=x23

  12. 118. g(x)=x+23

Practice PLUS

In Exercises 119122, use transformations of the graph of the greatest integer function, f(x)=int(x) to graph each function. (The graph of f(x)=int(x) is shown in Figure 1.39.)

  1. 119. g(x)=2 int (x+1)

  2. 120. g(x)=3 int (x1)

  3. 121. h(x)=int(x)+1

  4. 122. h(x)=int(x)1

In Exercises 123126, write a possible equation for the function whose graph is shown. Each graph shows a transformation of a common function.

  1. 123.

    A viewing window displays a concave down increasing curve that rises from (2, 0), passes through (5.5, 2), and (8, 2.5). All values are estimated.
  2. 124.

    A viewing window displays a curve that passes through (negative 6, 6), (negative 3, 3), (0, 2), (3.5, 0), and (6, negative 6). All values are estimated.
  3. 125.

    A viewing window displays an upward opening parabola that falls through (negative 8, 5), (negative 6, 0) with its vertex at (negative 3, negative 4) and rises through (2, 0), (5, 8). All values are estimated.
  4. 126.

    A viewing window displays a concave down increasing curve that rises from (2, 1), passes through (6, 3), and (9, 3.5). All values are estimated.

Application Exercises

  1. 127. The function f(x)=2.9x+20.1 models the median height, f(x), in inches, of boys who are x months of age. The graph of f is shown.

    A graph titled, boys height plots age in months versus median height in inches. The horizontal axis ranges from 0 to 60 in increments of 10, and the vertical axis ranges from 0 to 50 in increments of 10.

    Source: Laura Walther Nathanson, The Portable Pediatrician for Parents

    1. Describe how the graph can be obtained using transformations of the square root function f(x)=x.

    2. According to the model, what is the median height of boys who are 48 months, or four years, old? Use a calculator and round to the nearest tenth of an inch. The actual median height for boys at 48 months is 40.8 inches. How well does the model describe the actual height?

    3. Use the model to find the average rate of change, in inches per month, between birth and 10 months. Round to the nearest tenth.

    4. Use the model to find the average rate of change, in inches per month, between 50 and 60 months. Round to the nearest tenth. How does this compare with your answer in part (c)? How is this difference shown by the graph?

  2. 128. The function f(x)=3.1x+19 models the median height, f(x), in inches, of girls who are x months of age. The graph of f is shown.

    A graph titled, girls height plots age in months versus median height in inches. The horizontal axis ranges from 0 to 60 in increments of 10, and the vertical axis ranges from 0 to 50 in increments of 10.

    Source: Laura Walther Nathanson, The Portable Pediatrician for Parents

    1. Describe how the graph can be obtained using transformations of the square root function f(x)=x.

    2. According to the model, what is the median height of girls who are 48 months, or 4 years, old? Use a calculator and round to the nearest tenth of an inch. The actual median height for girls at 48 months is 40.2 inches. How well does the model describe the actual height?

    3. Use the model to find the average rate of change, in inches per month, between birth and 10 months. Round to the nearest tenth.

    4. Use the model to find the average rate of change, in inches per month, between 50 and 60 months. Round to the nearest tenth. How does this compare with your answer in part (c)? How is this difference shown by the graph?

Explaining the Concepts

  1. 129. What must be done to a function’s equation so that its graph is shifted vertically upward?

  2. 130. What must be done to a function’s equation so that its graph is shifted horizontally to the right?

  3. 131. What must be done to a function’s equation so that its graph is reflected about the x-axis?

  4. 132. What must be done to a function’s equation so that its graph is reflected about the y-axis?

  5. 133. What must be done to a function’s equation so that its graph is stretched vertically?

  6. 134. What must be done to a function’s equation so that its graph is shrunk horizontally?

Technology Exercises

  1. 135.

    1. Use a graphing utility to graph f(x)=x2+1.

    2. Graph f(x)=x2+1, g(x)=f(2x), h(x)=f(3x), and k(x)=f(4x) in the same viewing rectangle.

    3. Describe the relationship among the graphs of f, g, h, and k, with emphasis on different values of x for points on all four graphs that give the same y-coordinate.

    4. Generalize by describing the relationship between the graph of f and the graph of g, where g(x)=f(cx) for c>1.

    5. Try out your generalization by sketching the graphs of f(cx) for c=1, c=2, c=3, and c=4 for a function of your choice.

  2. 136.

    1. Use a graphing utility to graph f(x)=x2+1.

    2. Graph f(x)=x2+1, g(x)=f(12 x), and h(x)=f(14 x) in the same viewing rectangle.

    3. Describe the relationship among the graphs of f, g, and h, with emphasis on different values of x for points on all three graphs that give the same y-coordinate.

    4. Generalize by describing the relationship between the graph of f and the graph of g, where g(x)=f(cx) for 0<c<1.

    5. Try out your generalization by sketching the graphs of f(cx) for c=1, and c=12, and c=14 for a function of your choice.

Critical Thinking Exercises

Make Sense? During the winter, you program your home thermostat so that at midnight, the temperature is 55°. This temperature is maintained until 6 a.m. Then the house begins to warm up so that by 9 a.m. the temperature is 65°. At 6 p.m. the house begins to cool. By 9 p.m., the temperature is again 55°. The graph illustrates home temperature, f(t), as a function of hours after midnight, t.

A graph titled, home temperature as a function of time plots hours after midnight versus temperature in degree Fahrenheit.

In Exercises 137140, determine whether each statement makes sense or does not make sense, and explain your reasoning. If the statement makes sense, graph the new function on the domain [0, 24]. If the statement does not make sense, correct the function in the statement and graph the corrected function on the domain [0, 24].

  1. 137. I decided to keep the house 5° warmer than before, so I reprogrammed the thermostat to y=f(t)+5.

  2. 138. I decided to keep the house 5° cooler than before, so I reprogrammed the thermostat to y=f(t)5.

  3. 139. I decided to change the heating schedule to start one hour earlier than before, so I reprogrammed the thermostat to y=f(t1).

  4. 140. I decided to change the heating schedule to start one hour later than before, so I reprogrammed the thermostat to y=f(t+1).

In Exercises 141144, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 141. If f(x)=|x| and g(x)=|x+3|+3, then the graph of g is a translation of the graph of f 3 units to the right and 3 units upward.

  2. 142. If f(x)=x and g(x)=x, then f and g have identical graphs.

  3. 143. If f(x)=x2 and g(x)=5(x22), then the graph of g can be obtained from the graph of f by stretching f 5 units followed by a downward shift of 2 units.

  4. 144. If f(x)=x3 and g(x)=(x3)34, then the graph of g can be obtained from the graph of f by moving f 3 units to the right, reflecting about the x-axis, and then moving the resulting graph down 4 units.

In Exercises 145148, functions f and g are graphed in the same rectangular coordinate system. If g is obtained from f through a sequence of transformations, find an equation for g.

  1. 145.

    The image shows the graph plots two parabolas.
  2. 146.

    The image shows a graph that plots two v shaped curves.
  3. 147.

    A graph plots two curves.
  4. 148.

    The image shows a graph that plots a semicircle and a curve.

For Exercises 149152, assume that (a, b) is a point on the graph of f. What is the corresponding point on the graph of each of the following functions?

  1. 149. y=f(x)

  2. 150. y=2f(x)

  3. 151. y=f(x3)

  4. 152. y=f(x)3

Retaining the Concepts

  1. 153. The length of a rectangle exceeds the width by 13 yards. If the perimeter of the rectangle is 82 yards, what are its dimensions?

    (Section P.8, Example 6)

  2. 154. Solve: x+104=x.

    (Section P.7, Example 12)

  3. 155. if f(x)=x2+3x+2, find f(x+h)f(x)h, h0, and simplify.

    (Section 1.3, Example 8)

Preview Exercises

Exercises 156158 will help you prepare for the material covered in the next section.

In Exercises 156157, perform the indicated operation or operations.

  1. 156. (2x1)(x2+x2)

  2. 157. (f(x))22f(x)+6, where f(x)=3x4

  3. 158. Simplify: 23x1.

1.6: Exercise Set

1.6 Exercise Set

Practice Exercises

In Exercises 116, use the graph of y=f(x) to graph each function g.

The graph of y = f of x is a line segment that starts at (negative 2, 2), passes through (0, 2), and ends at (2, 2).
  1. 1. g(x)=f(x)+1

  2. 2. g(x)=f(x)1

  3. 3. g(x)=f(x+1)

  4. 4. g(x)=f(x1)

  5. 5. g(x)=f(x1)2

  6. 6. g(x)=f(x+1)+2

  7. 7. g(x)=f(x)

  8. 8. g(x)=f(x)

  9. 9. g(x)=f(x)+3

  10. 10. g(x)=f(x)+3

  11. 11. g(x)=12 f(x)

  12. 12. g(x)=2f(x)

  13. 13. g(x)=f(12 x)

  14. 14. g(x)=f(2x)

  15. 15. g(x)=f(12 x)+1

  16. 16. g(x)=f(2x)1

In Exercises 1732, use the graph of y=f(x) to graph each function g.

The graph of y = f of x is a curve that starts at (negative 4, 0), passes through (negative 2, negative 2), (0, 0), (2, 2), and ends at (4, 0).
  1. 17. g(x)=f(x)1

  2. 18. g(x)=f(x)+1

  3. 19. g(x)=f(x1)

  4. 20. g(x)=f(x+1)

  5. 21. g(x)=f(x1)+2

  6. 22. g(x)=f(x+1)2

  7. 23. g(x)=f(x)

  8. 24. g(x)=f(x)

  9. 25. g(x)=f(x)+1

  10. 26. g(x)=f(x)+1

  11. 27. g(x)=2f(x)

  12. 28. g(x)=12 f(x)

  13. 29. g(x)=f(2x)

  14. 30. g(x)=f(12 x)

  15. 31. g(x)=2f(x+2)+1

  16. 32. g(x)=2f(x+2)1

In Exercises 3344, use the graph of y=f(x) to graph each function g.

A graph plots two lines for y = f of x. The first line falls from a closed point (0, 0) to an open point (4, negative 2). The second line falls from a closed point (negative 4, 0) to an open point (0, negative 2).
  1. 33. g(x)=f(x)+2

  2. 34. g(x)=f(x)2

  3. 35. g(x)=f(x+2)

  4. 36. g(x)=f(x2)

  5. 37. g(x)=f(x+2)

  6. 38. g(x)=f(x2)

  7. 39. g(x)= 12f(x+2)

  8. 40. g(x)= 12f(x2)

  9. 41. g(x)= 12f(x+2)2

  10. 42. g(x)= 12f(x2)+2

  11. 43. g(x)=12f(2x)

  12. 44. g(x)=2f(12 x)

In Exercises 4552, use the graph of y=f(x) to graph each function g.

A graph of y = f of x is a continuous line that starts from (negative 4, negative 2) passes through (negative 2, 0), (0, 2) (2, 2) and ends at (4, 0).
  1. 45. g(x)=f(x1)1

  2. 46. g(x)=f(x+1)+1

  3. 47. g(x)=f(x1)+1

  4. 48. g(x)=f(x+1)1

  5. 49. g(x)=2f(12 x)

  6. 50. g(x)=12f(2x)

  7. 51. g(x)=12f(x+1)

  8. 52. g(x)=2f(x1)

In Exercises 5366, begin by graphing the standard quadratic function, f(x)=x2. Then use transformations of this graph to graph the given function.

  1. 53. g(x)=x22

  2. 54. g(x)=x21

  3. 55. g(x)=(x2)2

  4. 56. g(x)=(x1)2

  5. 57. h(x)=(x2)2

  6. 58. h(x)=(x1)2

  7. 59. h(x)=(x2)2+1

  8. 60. h(x)=(x1)2+2

  9. 61. g(x)=2(x2)2

  10. 62. g(x)=12 (x1)2

  11. 63. h(x)=2(x2)21

  12. 64. h(x)=12 (x1)21

  13. 65. h(x)=2(x+1)2+1

  14. 66. h(x)=2(x+2)2+1

In Exercises 6780, begin by graphing the square root function, f(x)=x. Then use transformations of this graph to graph the given function.

  1. 67. g(x)=x+2

  2. 68. g(x)=x+1

  3. 69. g(x)=x+2

  4. 70. g(x)=x+1

  5. 71. h(x)=x+2

  6. 72. h(x)=x+1

  7. 73. h(x)=x+2

  8. 74. h(x)=x+1

  9. 75. g(x)=12x+2

  10. 76. g(x)=2x+1

  11. 77. h(x)=x+22

  12. 78. h(x)=x+11

  13. 79. g(x)=2x+22

  14. 80. g(x)=2x+11

In Exercises 8194, begin by graphing the absolute value function, f(x)=|x|. Then use transformations of this graph to graph the given function.

  1. 81. g(x)=|x|+4

  2. 82. g(x)=|x|+3

  3. 83. g(x)=|x+4|

  4. 84. g(x)=|x+3|

  5. 85. h(x)=|x+4|2

  6. 86. h(x)=|x+3|2

  7. 87. h(x)=|x+4|

  8. 88. h(x)=|x+3|

  9. 89. g(x)=|x+4|+1

  10. 90. g(x)=|x+4|+2

  11. 91. h(x)=2|x+4|

  12. 92. h(x)=2|x+3|

  13. 93. g(x)=2|x+4|+1

  14. 94. g(x)=2|x+3|+2

In Exercises 95106, begin by graphing the standard cubic function, f(x)=x3. Then use transformations of this graph to graph the given function.

  1. 95. g(x)=x33

  2. 96. g(x)=x32

  3. 97. g(x)=(x3)3

  4. 98. g(x)=(x2)3

  5. 99. h(x)=x3

  6. 100. h(x)=(x2)3

  7. 101. h(x)=12 x3

  8. 102. h(x)=14 x3

  9. 103. r(x)=(x3)3+2

  10. 104. r(x)=(x2)3+1

  11. 105. h(x)=12 (x3)32

  12. 106. h(x)=12 (x2)31

In Exercises 107118, begin by graphing the cube root function, f(x)=x3. Then use transformations of this graph to graph the given function.

  1. 107. g(x)=x3+2

  2. 108. g(x)=x32

  3. 109. g(x)=x+23

  4. 110. g(x)=x23

  5. 111. h(x)=12x+23

  6. 112. h(x)=12x23

  7. 113. r(x)=12x+232

  8. 114. r(x)=12x23+2

  9. 115. h(x)=x+23

  10. 116. h(x)=x23

  11. 117. g(x)=x23

  12. 118. g(x)=x+23

Practice PLUS

In Exercises 119122, use transformations of the graph of the greatest integer function, f(x)=int(x) to graph each function. (The graph of f(x)=int(x) is shown in Figure 1.39.)

  1. 119. g(x)=2 int (x+1)

  2. 120. g(x)=3 int (x1)

  3. 121. h(x)=int(x)+1

  4. 122. h(x)=int(x)1

In Exercises 123126, write a possible equation for the function whose graph is shown. Each graph shows a transformation of a common function.

  1. 123.

    A viewing window displays a concave down increasing curve that rises from (2, 0), passes through (5.5, 2), and (8, 2.5). All values are estimated.
  2. 124.

    A viewing window displays a curve that passes through (negative 6, 6), (negative 3, 3), (0, 2), (3.5, 0), and (6, negative 6). All values are estimated.
  3. 125.

    A viewing window displays an upward opening parabola that falls through (negative 8, 5), (negative 6, 0) with its vertex at (negative 3, negative 4) and rises through (2, 0), (5, 8). All values are estimated.
  4. 126.

    A viewing window displays a concave down increasing curve that rises from (2, 1), passes through (6, 3), and (9, 3.5). All values are estimated.

Application Exercises

  1. 127. The function f(x)=2.9x+20.1 models the median height, f(x), in inches, of boys who are x months of age. The graph of f is shown.

    A graph titled, boys height plots age in months versus median height in inches. The horizontal axis ranges from 0 to 60 in increments of 10, and the vertical axis ranges from 0 to 50 in increments of 10.

    Source: Laura Walther Nathanson, The Portable Pediatrician for Parents

    1. Describe how the graph can be obtained using transformations of the square root function f(x)=x.

    2. According to the model, what is the median height of boys who are 48 months, or four years, old? Use a calculator and round to the nearest tenth of an inch. The actual median height for boys at 48 months is 40.8 inches. How well does the model describe the actual height?

    3. Use the model to find the average rate of change, in inches per month, between birth and 10 months. Round to the nearest tenth.

    4. Use the model to find the average rate of change, in inches per month, between 50 and 60 months. Round to the nearest tenth. How does this compare with your answer in part (c)? How is this difference shown by the graph?

  2. 128. The function f(x)=3.1x+19 models the median height, f(x), in inches, of girls who are x months of age. The graph of f is shown.

    A graph titled, girls height plots age in months versus median height in inches. The horizontal axis ranges from 0 to 60 in increments of 10, and the vertical axis ranges from 0 to 50 in increments of 10.

    Source: Laura Walther Nathanson, The Portable Pediatrician for Parents

    1. Describe how the graph can be obtained using transformations of the square root function f(x)=x.

    2. According to the model, what is the median height of girls who are 48 months, or 4 years, old? Use a calculator and round to the nearest tenth of an inch. The actual median height for girls at 48 months is 40.2 inches. How well does the model describe the actual height?

    3. Use the model to find the average rate of change, in inches per month, between birth and 10 months. Round to the nearest tenth.

    4. Use the model to find the average rate of change, in inches per month, between 50 and 60 months. Round to the nearest tenth. How does this compare with your answer in part (c)? How is this difference shown by the graph?

Explaining the Concepts

  1. 129. What must be done to a function’s equation so that its graph is shifted vertically upward?

  2. 130. What must be done to a function’s equation so that its graph is shifted horizontally to the right?

  3. 131. What must be done to a function’s equation so that its graph is reflected about the x-axis?

  4. 132. What must be done to a function’s equation so that its graph is reflected about the y-axis?

  5. 133. What must be done to a function’s equation so that its graph is stretched vertically?

  6. 134. What must be done to a function’s equation so that its graph is shrunk horizontally?

Technology Exercises

  1. 135.

    1. Use a graphing utility to graph f(x)=x2+1.

    2. Graph f(x)=x2+1, g(x)=f(2x), h(x)=f(3x), and k(x)=f(4x) in the same viewing rectangle.

    3. Describe the relationship among the graphs of f, g, h, and k, with emphasis on different values of x for points on all four graphs that give the same y-coordinate.

    4. Generalize by describing the relationship between the graph of f and the graph of g, where g(x)=f(cx) for c>1.

    5. Try out your generalization by sketching the graphs of f(cx) for c=1, c=2, c=3, and c=4 for a function of your choice.

  2. 136.

    1. Use a graphing utility to graph f(x)=x2+1.

    2. Graph f(x)=x2+1, g(x)=f(12 x), and h(x)=f(14 x) in the same viewing rectangle.

    3. Describe the relationship among the graphs of f, g, and h, with emphasis on different values of x for points on all three graphs that give the same y-coordinate.

    4. Generalize by describing the relationship between the graph of f and the graph of g, where g(x)=f(cx) for 0<c<1.

    5. Try out your generalization by sketching the graphs of f(cx) for c=1, and c=12, and c=14 for a function of your choice.

Critical Thinking Exercises

Make Sense? During the winter, you program your home thermostat so that at midnight, the temperature is 55°. This temperature is maintained until 6 a.m. Then the house begins to warm up so that by 9 a.m. the temperature is 65°. At 6 p.m. the house begins to cool. By 9 p.m., the temperature is again 55°. The graph illustrates home temperature, f(t), as a function of hours after midnight, t.

A graph titled, home temperature as a function of time plots hours after midnight versus temperature in degree Fahrenheit.

In Exercises 137140, determine whether each statement makes sense or does not make sense, and explain your reasoning. If the statement makes sense, graph the new function on the domain [0, 24]. If the statement does not make sense, correct the function in the statement and graph the corrected function on the domain [0, 24].

  1. 137. I decided to keep the house 5° warmer than before, so I reprogrammed the thermostat to y=f(t)+5.

  2. 138. I decided to keep the house 5° cooler than before, so I reprogrammed the thermostat to y=f(t)5.

  3. 139. I decided to change the heating schedule to start one hour earlier than before, so I reprogrammed the thermostat to y=f(t1).

  4. 140. I decided to change the heating schedule to start one hour later than before, so I reprogrammed the thermostat to y=f(t+1).

In Exercises 141144, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 141. If f(x)=|x| and g(x)=|x+3|+3, then the graph of g is a translation of the graph of f 3 units to the right and 3 units upward.

  2. 142. If f(x)=x and g(x)=x, then f and g have identical graphs.

  3. 143. If f(x)=x2 and g(x)=5(x22), then the graph of g can be obtained from the graph of f by stretching f 5 units followed by a downward shift of 2 units.

  4. 144. If f(x)=x3 and g(x)=(x3)34, then the graph of g can be obtained from the graph of f by moving f 3 units to the right, reflecting about the x-axis, and then moving the resulting graph down 4 units.

In Exercises 145148, functions f and g are graphed in the same rectangular coordinate system. If g is obtained from f through a sequence of transformations, find an equation for g.

  1. 145.

    The image shows the graph plots two parabolas.
  2. 146.

    The image shows a graph that plots two v shaped curves.
  3. 147.

    A graph plots two curves.
  4. 148.

    The image shows a graph that plots a semicircle and a curve.

For Exercises 149152, assume that (a, b) is a point on the graph of f. What is the corresponding point on the graph of each of the following functions?

  1. 149. y=f(x)

  2. 150. y=2f(x)

  3. 151. y=f(x3)

  4. 152. y=f(x)3

Retaining the Concepts

  1. 153. The length of a rectangle exceeds the width by 13 yards. If the perimeter of the rectangle is 82 yards, what are its dimensions?

    (Section P.8, Example 6)

  2. 154. Solve: x+104=x.

    (Section P.7, Example 12)

  3. 155. if f(x)=x2+3x+2, find f(x+h)f(x)h, h0, and simplify.

    (Section 1.3, Example 8)

Preview Exercises

Exercises 156158 will help you prepare for the material covered in the next section.

In Exercises 156157, perform the indicated operation or operations.

  1. 156. (2x1)(x2+x2)

  2. 157. (f(x))22f(x)+6, where f(x)=3x4

  3. 158. Simplify: 23x1.

Section 1.7: Combinations of Functions; Composite Functions

Section 1.7 Combinations of Functions; Composite Functions

Learning Objectives

What You’ll Learn

  1. 1 Find the domain of a function.

  2. 2 Combine functions using the algebra of functions, specifying domains.

  3. 3 Form composite functions.

  4. 4 Determine domains for composite functions.

  5. 5 Write functions as compositions.

We’re born. We die. Figure 1.66 quantifies these statements by showing the number of births and deaths in the United States for nine selected years.

Figure 1.66

A bar graph titled, number of births and deaths in the United States plots number, in thousands versus year. The horizontal axis ranges from 20 00 to 20 16, in increments of 2 years. The vertical axis ranges from 2000 to 4400, in increments of 300.

Source: U.S. Department of Health and Human Services

Figure 1.66 Full Alternative Text

In this section, we look at these data from the perspective of functions. By considering the yearly change in the U.S. population, you will see that functions can be subtracted using procedures that will remind you of combining algebraic expressions.

Objective 1: Find the domain of a function

The Domain of a Function

  1. Objective 1 Find the domain of a function.

Watch Video

We begin with two functions that model the data in Figure 1.66.

B of x = negative 2.6 x squared + 33 x + 4043, labeled, number of births comma B of x comma in thousands comma x years after 2000.
1.7-272 Full Alternative Text

The data in Figure 1.66 show even-numbered years from 2000 through 2016. Because x represents the number of years after 2000,

Domain ofB={0, 2, 4, 6,, 16}

and

Domain ofD={0, 2, 4, 6,, 16}.

Functions that model data often have their domains explicitly given with the function’s equation. However, for most functions, only an equation is given and the domain is not specified. In cases like this, the domain of a function f is the largest set of real numbers for which the value of f(x) is a real number. For example, consider the function

f(x)=1x3.

Because division by 0 is undefined, the denominator, x3, cannot be 0. Thus, x cannot equal 3. The domain of the function consists of all real numbers other than 3, represented by

Domain off={x|x is a real number andx3}.

Using interval notation,

The domain of f = the union of (negative infinity comma 3) and (3 comma infinity). (negative infinity, 3) is labeled, all real numbers less than 3. Union symbol is labeled, or and (3, infinity) is labeled, all real numbers greater than 3.

Now consider a function involving a square root:

g(x)=x3.

Because only nonnegative numbers have square roots that are real numbers, the expression under the square root sign, x3, must be nonnegative. We can use inspection to see that x30 if x3. The domain of g consists of all real numbers that are greater than or equal to 3:

Domain of g={x|x3}or [3, ).

Finding a Function’s Domain

If a function f does not model data or verbal conditions, its domain is the largest set of real numbers for which the value of f(x) is a real number. Exclude from a function’s domain real numbers that cause division by zero and real numbers that result in an even root, such as a square root, of a negative number.

Example 1 Finding the Domain of a Function

Find the domain of each function:

  1. f(x)=x27x

  2. g(x)=3x+2x22x3

  3. h(x)=3x+12

  4. j(x)=3x+2142x.

Solution

The domain is the set of all real numbers, (, ), unless x appears in a denominator or in an even root, such as a square root.

  1. The function f(x)=x27x contains neither division nor a square root. For every real number, x, the algebraic expression x27x represents a real number. Thus, the domain of f is the set of all real numbers.

    Domain of f=(, )
  2. The function g(x)=3x+2x22x3 contains division. Because division by 0 is undefined, we must exclude from the domain the values of x that cause the denominator, x22x3, to be 0. We can identify these values by setting x22x3 equal to 0.

    x22x3=0Set the functions denominator equal to 0.(x+1)(x3)=0Factor.x+1=0       or     x3=0Set each factor equal to 0.x=1x=3Solve the resulting equations.

    We must exclude 1 and 3 from the domain of g(x)=3x+2x22x3.

    Domain of g=(, 1)(1, 3)(3, )
  3. The function h(x)=3x+12 contains an even root. Because only nonnegative numbers have real square roots, the quantity under the radical sign, 3x+12, must be greater than or equal to 0.

    3x+120Set the functions radicand greater than or equal to 0.3x12Subtract 12 from both sides.x4Divide both sides by 3. Division by a positive number preserves the sense of the inequality.

    The domain of h consists of all real numbers greater than or equal to 4.

    Domain of h=[4, )

    The domain is highlighted on the x-axis in Figure 1.67.

    Figure 1.67

    A viewing window displays a curve for h of x = square root of start expression 3 x + 12 end expression
  4. The function j(x)=3x+2142x contains both an even root and division. Because only nonnegative numbers have real square roots, the quantity under the radical sign, 142x, must be greater than or equal to 0. But wait, there’s more! Because division by 0 is undefined, 142x cannot equal 0. Thus, 142x must be strictly greater than 0.

    14-2x>0Set the functions radicand greater than 0.-2x>-14Subtract 14 from both sides.x<7Divide both sides by-2. Division by a negativenumber reverses the direction of the inequality.

    The domain of j consists of all real numbers less than 7.

    Domain of j=(, 7)

Check Point 1

  • Find the domain of each function:

    1. f(x)=x2+3x17

    2. g(x)=5xx249

    3. h(x)=9x27

    4. j(x)=5x243x.

Objective 1: Find the domain of a function

The Domain of a Function

  1. Objective 1 Find the domain of a function.

Watch Video

We begin with two functions that model the data in Figure 1.66.

B of x = negative 2.6 x squared + 33 x + 4043, labeled, number of births comma B of x comma in thousands comma x years after 2000.
1.7-272 Full Alternative Text

The data in Figure 1.66 show even-numbered years from 2000 through 2016. Because x represents the number of years after 2000,

Domain ofB={0, 2, 4, 6,, 16}

and

Domain ofD={0, 2, 4, 6,, 16}.

Functions that model data often have their domains explicitly given with the function’s equation. However, for most functions, only an equation is given and the domain is not specified. In cases like this, the domain of a function f is the largest set of real numbers for which the value of f(x) is a real number. For example, consider the function

f(x)=1x3.

Because division by 0 is undefined, the denominator, x3, cannot be 0. Thus, x cannot equal 3. The domain of the function consists of all real numbers other than 3, represented by

Domain off={x|x is a real number andx3}.

Using interval notation,

The domain of f = the union of (negative infinity comma 3) and (3 comma infinity). (negative infinity, 3) is labeled, all real numbers less than 3. Union symbol is labeled, or and (3, infinity) is labeled, all real numbers greater than 3.

Now consider a function involving a square root:

g(x)=x3.

Because only nonnegative numbers have square roots that are real numbers, the expression under the square root sign, x3, must be nonnegative. We can use inspection to see that x30 if x3. The domain of g consists of all real numbers that are greater than or equal to 3:

Domain of g={x|x3}or [3, ).

Finding a Function’s Domain

If a function f does not model data or verbal conditions, its domain is the largest set of real numbers for which the value of f(x) is a real number. Exclude from a function’s domain real numbers that cause division by zero and real numbers that result in an even root, such as a square root, of a negative number.

Example 1 Finding the Domain of a Function

Find the domain of each function:

  1. f(x)=x27x

  2. g(x)=3x+2x22x3

  3. h(x)=3x+12

  4. j(x)=3x+2142x.

Solution

The domain is the set of all real numbers, (, ), unless x appears in a denominator or in an even root, such as a square root.

  1. The function f(x)=x27x contains neither division nor a square root. For every real number, x, the algebraic expression x27x represents a real number. Thus, the domain of f is the set of all real numbers.

    Domain of f=(, )
  2. The function g(x)=3x+2x22x3 contains division. Because division by 0 is undefined, we must exclude from the domain the values of x that cause the denominator, x22x3, to be 0. We can identify these values by setting x22x3 equal to 0.

    x22x3=0Set the functions denominator equal to 0.(x+1)(x3)=0Factor.x+1=0       or     x3=0Set each factor equal to 0.x=1x=3Solve the resulting equations.

    We must exclude 1 and 3 from the domain of g(x)=3x+2x22x3.

    Domain of g=(, 1)(1, 3)(3, )
  3. The function h(x)=3x+12 contains an even root. Because only nonnegative numbers have real square roots, the quantity under the radical sign, 3x+12, must be greater than or equal to 0.

    3x+120Set the functions radicand greater than or equal to 0.3x12Subtract 12 from both sides.x4Divide both sides by 3. Division by a positive number preserves the sense of the inequality.

    The domain of h consists of all real numbers greater than or equal to 4.

    Domain of h=[4, )

    The domain is highlighted on the x-axis in Figure 1.67.

    Figure 1.67

    A viewing window displays a curve for h of x = square root of start expression 3 x + 12 end expression
  4. The function j(x)=3x+2142x contains both an even root and division. Because only nonnegative numbers have real square roots, the quantity under the radical sign, 142x, must be greater than or equal to 0. But wait, there’s more! Because division by 0 is undefined, 142x cannot equal 0. Thus, 142x must be strictly greater than 0.

    14-2x>0Set the functions radicand greater than 0.-2x>-14Subtract 14 from both sides.x<7Divide both sides by-2. Division by a negativenumber reverses the direction of the inequality.

    The domain of j consists of all real numbers less than 7.

    Domain of j=(, 7)

Check Point 1

  • Find the domain of each function:

    1. f(x)=x2+3x17

    2. g(x)=5xx249

    3. h(x)=9x27

    4. j(x)=5x243x.

Objective 2: Combine functions using the algebra of functions, specifying domains

The Algebra of Functions

  1. Objective 2 Combine functions using the algebra of functions, specifying domains.

Watch Video

We can combine functions using addition, subtraction, multiplication, and division by performing operations with the algebraic expressions that appear on the right side of the equations. For example, the functions f(x)=2x and g(x)=x1 can be combined to form the sum, difference, product, and quotient of f and g. Here’s how it’s done:

The image shows the solutions for the sum, difference, product, and quotient of f and g.
1.7-275 Full Alternative Text

The domain for each of these functions consists of all real numbers that are common to the domains of f and g. Using Df to represent the domain of f and Dg to represent the domain of g, the domain for each function is DfDg. In the case of the quotient function f(x)g(x), we must remember not to divide by 0, so we add the further restriction that g(x)0.

The Algebra of Functions: Sum, Difference, Product, and Quotient of Functions

Let f and g be two functions. The sum f+g, the difference fg, the product fg, and the quotient fg are functions whose domains are the set of all real numbers common to the domains of f and g(DfDg), defined as follows:

1. Sum: (f+g)(x)=f(x)+g(x)
2. Difference: (fg)(x)=f(x)g(x)
3. Product: (fg)(x)=f(x)g(x)
4. Quotient: (fg)(x)=f(x)g(x), provided g(x)0.

Example 2 Combining Functions

Let f(x)=2x1 and g(x)=x2+x2. Find each of the following functions:

  1. (f+g)(x)

  2. (fg)(x)

  3. (fg)(x)

  4. (fg)(x).

Determine the domain for each function.

Solution

  1.  

    (f+g)(x)=f(x)+g(x)This is the definition of the sum f+g.(f+g)(x)=(2x1)+(x2+x2)Substitute the given functions.(f+g)(x)=x2+3x3Remove parentheses and combine like terms.
  2.  

    (fg)(x)=f(x)g(x)This is the definition of the difference fg.(f+g)(x)=(2x1)(x2+x2)Substitute the given functions.(f+g)(x)=2x1x2x+2Remove parentheses and change the sign ofeach term in the second set of parentheses.(f+g)(x)=x2+x+1Combine like terms and arrange terms indescending powers of x.

  3.  

    (fg)(x)=f(x)g(x)This is the definition of theproduct fg.=(2x1)(x2+x+2)Substitute the given functions.=2x(x2+x2)1(x2+x2)Multiply each term in thesecond factor by 2x and 1,respectively.=2x3+2x24xx2x+2Use the distributive property.=2x3+(2x2x2)+(4xx)+2Rearrange terms so that liketerms are adjacent.=2x3+x25x+2Combine like terms.
  4.  

    (fg)(x)=f(x)g(x)This is the definition of the quotient fg.=2x1x2+x2Substitute the given functions. This rational expressioncannot be simplified.

Because the equations for f and g do not involve division or contain even roots, the domain of both f and g is the set of all real numbers. Thus, the domain of f+g, fg, and fg is the set of all real numbers, (, ).

The function fg contains division. We must exclude from its domain values of x that cause the denominator, x2+x2, to be 0. Let’s identify these values.

x2+x2=0Set the denominator of fg equal to 0.(x+2)(x1)=0Factor.x+2=0orx1=0Set each factor equal to 0.x=2x=1Solve the resulting equations.

We must exclude 2 and 1 from the domain of fg.

Domain offg=(, 2)(2, 1)(1, )

Check Point 2

  • Let f(x)=x5 and g(x)=x21. Find each of the following functions:

    1. (f+g)(x)

    2. (fg)(x)

    3. (fg)(x)

    4. (fg)(x).

    Determine the domain for each function.

Example 3 Adding Functions and Determining the Domain

Let f(x)=x+3 and g(x)=x2. Find each of the following:

  1. (f+g)(x)

  2. the domain of f+g.

Solution

  1. (f+g)(x)=f(x)+g(x)=x++x2

  2. The domain of f+g is the set of all real numbers that are common to the domain of f and the domain of g. Thus, we must find the domains of f and g before finding their intersection.

f of x = square root of start expression x + 3 end expression. g of x = square root of start expression x minus 2 end expression.

Now, we can use a number line to determine DfDg, the domain of f+g. Figure 1.68 shows the domain of f, [3, ), in blue and the domain of g, [2, ), in red. Can you see that all real numbers greater than or equal to 2 are common to both domains? This is shown in purple on the number line. Thus, the domain of f+g is [2, ).

Figure 1.68 Finding the domain of the sum f+g

A number line determines the domain of f, the domain of g, and the domain of f + g.

Check Point 3

  • Let f(x)=x3 and g(x)=x+1. Find each of the following:

    1. (f+g)(x)

    2. the domain of f+g.

Example 4 Applying the Algebra of Functions

We opened the section with functions that model the number of births and deaths in the United States for selected years from 2000 through 2016:

B of x = negative 2.6 x squared + 33 x + 4043, labeled, number of births comma B of x comma in thousands comma x years after 2000.
  1. Write a function that models the change in U.S. population, B(x)D(x), for each year from 2000 through 2016.

  2. Use the function from part (a) to find the change in U.S. population in 2014.

  3. Does the result in part (b) overestimate or underestimate the actual population change in 2014 obtained from the data in Figure 1.66? By how much?

Solution

  1. The change in population is the number of births minus the number of deaths. Thus, we will find the difference function, BD.

    (BD)(x)=B(x)D(x)=(2.6x2+33x+4043)(2x213x+2428)Substitute the given functions.=2.6x2+33x+4043+(2x3)+13x2428Remove parentheses andchange the sign of each term inthe second set of parentheses.=(2.6x22x2)+(33x+13x)+(40432428)Group like terms.=4.6x2+46x+1615Combine like terms.

    The function

    (BD)(x)=4.6x2+46x+1615

    models the change in U.S. population, in thousands, x years after 2000.

  2. Because 2014 is 14 years after 2000, we substitute 14 for x in the difference function (BD)(x)

    (BD)(x)=4.6x2+46x+1615Use the difference function BD.(BD)(14)=4.6(14)2+46(14)+1615Substitute 14 for x.=4.6(196)+46(14)+1615Evaluate the exponential expression:142=196.=901.6+644+1615Perform the multiplications.=1357.4Add from left to right.

    We see that (BD)(14)=1357.4 The model indicates that there was a population increase of 1357.4 thousand, or approximately 1,357,400 people, in 2014.

  3. The data for 2014 in Figure 1.66 show 3988 thousand births and 2628 thousand deaths.

population change=birthsdeathspopulation change=39882628=1360

The actual population increase was 1360 thousand, or 1,360,000. Our model gave us an increase of 1357.4 thousand. Thus, the model underestimates the actual increase by 13601357.4, or 2.6 thousand people.

Check Point 4

  • Use the birth and death models from Example 4.

    1. Write a function that models the total number of births and deaths in the United States for the years from 2000 through 2016.

    2. Use the function from part (a) to find the total number of births and deaths in the United States in 2000.

    3. Does the result in part (b) overestimate or underestimate the actual number of total births and deaths in 2000 obtained from the data in Figure 1.66? By how much?

Objective 2: Combine functions using the algebra of functions, specifying domains

The Algebra of Functions

  1. Objective 2 Combine functions using the algebra of functions, specifying domains.

Watch Video

We can combine functions using addition, subtraction, multiplication, and division by performing operations with the algebraic expressions that appear on the right side of the equations. For example, the functions f(x)=2x and g(x)=x1 can be combined to form the sum, difference, product, and quotient of f and g. Here’s how it’s done:

The image shows the solutions for the sum, difference, product, and quotient of f and g.
1.7-275 Full Alternative Text

The domain for each of these functions consists of all real numbers that are common to the domains of f and g. Using Df to represent the domain of f and Dg to represent the domain of g, the domain for each function is DfDg. In the case of the quotient function f(x)g(x), we must remember not to divide by 0, so we add the further restriction that g(x)0.

The Algebra of Functions: Sum, Difference, Product, and Quotient of Functions

Let f and g be two functions. The sum f+g, the difference fg, the product fg, and the quotient fg are functions whose domains are the set of all real numbers common to the domains of f and g(DfDg), defined as follows:

1. Sum: (f+g)(x)=f(x)+g(x)
2. Difference: (fg)(x)=f(x)g(x)
3. Product: (fg)(x)=f(x)g(x)
4. Quotient: (fg)(x)=f(x)g(x), provided g(x)0.

Example 2 Combining Functions

Let f(x)=2x1 and g(x)=x2+x2. Find each of the following functions:

  1. (f+g)(x)

  2. (fg)(x)

  3. (fg)(x)

  4. (fg)(x).

Determine the domain for each function.

Solution

  1.  

    (f+g)(x)=f(x)+g(x)This is the definition of the sum f+g.(f+g)(x)=(2x1)+(x2+x2)Substitute the given functions.(f+g)(x)=x2+3x3Remove parentheses and combine like terms.
  2.  

    (fg)(x)=f(x)g(x)This is the definition of the difference fg.(f+g)(x)=(2x1)(x2+x2)Substitute the given functions.(f+g)(x)=2x1x2x+2Remove parentheses and change the sign ofeach term in the second set of parentheses.(f+g)(x)=x2+x+1Combine like terms and arrange terms indescending powers of x.

  3.  

    (fg)(x)=f(x)g(x)This is the definition of theproduct fg.=(2x1)(x2+x+2)Substitute the given functions.=2x(x2+x2)1(x2+x2)Multiply each term in thesecond factor by 2x and 1,respectively.=2x3+2x24xx2x+2Use the distributive property.=2x3+(2x2x2)+(4xx)+2Rearrange terms so that liketerms are adjacent.=2x3+x25x+2Combine like terms.
  4.  

    (fg)(x)=f(x)g(x)This is the definition of the quotient fg.=2x1x2+x2Substitute the given functions. This rational expressioncannot be simplified.

Because the equations for f and g do not involve division or contain even roots, the domain of both f and g is the set of all real numbers. Thus, the domain of f+g, fg, and fg is the set of all real numbers, (, ).

The function fg contains division. We must exclude from its domain values of x that cause the denominator, x2+x2, to be 0. Let’s identify these values.

x2+x2=0Set the denominator of fg equal to 0.(x+2)(x1)=0Factor.x+2=0orx1=0Set each factor equal to 0.x=2x=1Solve the resulting equations.

We must exclude 2 and 1 from the domain of fg.

Domain offg=(, 2)(2, 1)(1, )

Check Point 2

  • Let f(x)=x5 and g(x)=x21. Find each of the following functions:

    1. (f+g)(x)

    2. (fg)(x)

    3. (fg)(x)

    4. (fg)(x).

    Determine the domain for each function.

Example 3 Adding Functions and Determining the Domain

Let f(x)=x+3 and g(x)=x2. Find each of the following:

  1. (f+g)(x)

  2. the domain of f+g.

Solution

  1. (f+g)(x)=f(x)+g(x)=x++x2

  2. The domain of f+g is the set of all real numbers that are common to the domain of f and the domain of g. Thus, we must find the domains of f and g before finding their intersection.

f of x = square root of start expression x + 3 end expression. g of x = square root of start expression x minus 2 end expression.

Now, we can use a number line to determine DfDg, the domain of f+g. Figure 1.68 shows the domain of f, [3, ), in blue and the domain of g, [2, ), in red. Can you see that all real numbers greater than or equal to 2 are common to both domains? This is shown in purple on the number line. Thus, the domain of f+g is [2, ).

Figure 1.68 Finding the domain of the sum f+g

A number line determines the domain of f, the domain of g, and the domain of f + g.

Check Point 3

  • Let f(x)=x3 and g(x)=x+1. Find each of the following:

    1. (f+g)(x)

    2. the domain of f+g.

Example 4 Applying the Algebra of Functions

We opened the section with functions that model the number of births and deaths in the United States for selected years from 2000 through 2016:

B of x = negative 2.6 x squared + 33 x + 4043, labeled, number of births comma B of x comma in thousands comma x years after 2000.
  1. Write a function that models the change in U.S. population, B(x)D(x), for each year from 2000 through 2016.

  2. Use the function from part (a) to find the change in U.S. population in 2014.

  3. Does the result in part (b) overestimate or underestimate the actual population change in 2014 obtained from the data in Figure 1.66? By how much?

Solution

  1. The change in population is the number of births minus the number of deaths. Thus, we will find the difference function, BD.

    (BD)(x)=B(x)D(x)=(2.6x2+33x+4043)(2x213x+2428)Substitute the given functions.=2.6x2+33x+4043+(2x3)+13x2428Remove parentheses andchange the sign of each term inthe second set of parentheses.=(2.6x22x2)+(33x+13x)+(40432428)Group like terms.=4.6x2+46x+1615Combine like terms.

    The function

    (BD)(x)=4.6x2+46x+1615

    models the change in U.S. population, in thousands, x years after 2000.

  2. Because 2014 is 14 years after 2000, we substitute 14 for x in the difference function (BD)(x)

    (BD)(x)=4.6x2+46x+1615Use the difference function BD.(BD)(14)=4.6(14)2+46(14)+1615Substitute 14 for x.=4.6(196)+46(14)+1615Evaluate the exponential expression:142=196.=901.6+644+1615Perform the multiplications.=1357.4Add from left to right.

    We see that (BD)(14)=1357.4 The model indicates that there was a population increase of 1357.4 thousand, or approximately 1,357,400 people, in 2014.

  3. The data for 2014 in Figure 1.66 show 3988 thousand births and 2628 thousand deaths.

population change=birthsdeathspopulation change=39882628=1360

The actual population increase was 1360 thousand, or 1,360,000. Our model gave us an increase of 1357.4 thousand. Thus, the model underestimates the actual increase by 13601357.4, or 2.6 thousand people.

Check Point 4

  • Use the birth and death models from Example 4.

    1. Write a function that models the total number of births and deaths in the United States for the years from 2000 through 2016.

    2. Use the function from part (a) to find the total number of births and deaths in the United States in 2000.

    3. Does the result in part (b) overestimate or underestimate the actual number of total births and deaths in 2000 obtained from the data in Figure 1.66? By how much?

Objective 2: Combine functions using the algebra of functions, specifying domains

The Algebra of Functions

  1. Objective 2 Combine functions using the algebra of functions, specifying domains.

Watch Video

We can combine functions using addition, subtraction, multiplication, and division by performing operations with the algebraic expressions that appear on the right side of the equations. For example, the functions f(x)=2x and g(x)=x1 can be combined to form the sum, difference, product, and quotient of f and g. Here’s how it’s done:

The image shows the solutions for the sum, difference, product, and quotient of f and g.
1.7-275 Full Alternative Text

The domain for each of these functions consists of all real numbers that are common to the domains of f and g. Using Df to represent the domain of f and Dg to represent the domain of g, the domain for each function is DfDg. In the case of the quotient function f(x)g(x), we must remember not to divide by 0, so we add the further restriction that g(x)0.

The Algebra of Functions: Sum, Difference, Product, and Quotient of Functions

Let f and g be two functions. The sum f+g, the difference fg, the product fg, and the quotient fg are functions whose domains are the set of all real numbers common to the domains of f and g(DfDg), defined as follows:

1. Sum: (f+g)(x)=f(x)+g(x)
2. Difference: (fg)(x)=f(x)g(x)
3. Product: (fg)(x)=f(x)g(x)
4. Quotient: (fg)(x)=f(x)g(x), provided g(x)0.

Example 2 Combining Functions

Let f(x)=2x1 and g(x)=x2+x2. Find each of the following functions:

  1. (f+g)(x)

  2. (fg)(x)

  3. (fg)(x)

  4. (fg)(x).

Determine the domain for each function.

Solution

  1.  

    (f+g)(x)=f(x)+g(x)This is the definition of the sum f+g.(f+g)(x)=(2x1)+(x2+x2)Substitute the given functions.(f+g)(x)=x2+3x3Remove parentheses and combine like terms.
  2.  

    (fg)(x)=f(x)g(x)This is the definition of the difference fg.(f+g)(x)=(2x1)(x2+x2)Substitute the given functions.(f+g)(x)=2x1x2x+2Remove parentheses and change the sign ofeach term in the second set of parentheses.(f+g)(x)=x2+x+1Combine like terms and arrange terms indescending powers of x.

  3.  

    (fg)(x)=f(x)g(x)This is the definition of theproduct fg.=(2x1)(x2+x+2)Substitute the given functions.=2x(x2+x2)1(x2+x2)Multiply each term in thesecond factor by 2x and 1,respectively.=2x3+2x24xx2x+2Use the distributive property.=2x3+(2x2x2)+(4xx)+2Rearrange terms so that liketerms are adjacent.=2x3+x25x+2Combine like terms.
  4.  

    (fg)(x)=f(x)g(x)This is the definition of the quotient fg.=2x1x2+x2Substitute the given functions. This rational expressioncannot be simplified.

Because the equations for f and g do not involve division or contain even roots, the domain of both f and g is the set of all real numbers. Thus, the domain of f+g, fg, and fg is the set of all real numbers, (, ).

The function fg contains division. We must exclude from its domain values of x that cause the denominator, x2+x2, to be 0. Let’s identify these values.

x2+x2=0Set the denominator of fg equal to 0.(x+2)(x1)=0Factor.x+2=0orx1=0Set each factor equal to 0.x=2x=1Solve the resulting equations.

We must exclude 2 and 1 from the domain of fg.

Domain offg=(, 2)(2, 1)(1, )

Check Point 2

  • Let f(x)=x5 and g(x)=x21. Find each of the following functions:

    1. (f+g)(x)

    2. (fg)(x)

    3. (fg)(x)

    4. (fg)(x).

    Determine the domain for each function.

Example 3 Adding Functions and Determining the Domain

Let f(x)=x+3 and g(x)=x2. Find each of the following:

  1. (f+g)(x)

  2. the domain of f+g.

Solution

  1. (f+g)(x)=f(x)+g(x)=x++x2

  2. The domain of f+g is the set of all real numbers that are common to the domain of f and the domain of g. Thus, we must find the domains of f and g before finding their intersection.

f of x = square root of start expression x + 3 end expression. g of x = square root of start expression x minus 2 end expression.

Now, we can use a number line to determine DfDg, the domain of f+g. Figure 1.68 shows the domain of f, [3, ), in blue and the domain of g, [2, ), in red. Can you see that all real numbers greater than or equal to 2 are common to both domains? This is shown in purple on the number line. Thus, the domain of f+g is [2, ).

Figure 1.68 Finding the domain of the sum f+g

A number line determines the domain of f, the domain of g, and the domain of f + g.

Check Point 3

  • Let f(x)=x3 and g(x)=x+1. Find each of the following:

    1. (f+g)(x)

    2. the domain of f+g.

Example 4 Applying the Algebra of Functions

We opened the section with functions that model the number of births and deaths in the United States for selected years from 2000 through 2016:

B of x = negative 2.6 x squared + 33 x + 4043, labeled, number of births comma B of x comma in thousands comma x years after 2000.
  1. Write a function that models the change in U.S. population, B(x)D(x), for each year from 2000 through 2016.

  2. Use the function from part (a) to find the change in U.S. population in 2014.

  3. Does the result in part (b) overestimate or underestimate the actual population change in 2014 obtained from the data in Figure 1.66? By how much?

Solution

  1. The change in population is the number of births minus the number of deaths. Thus, we will find the difference function, BD.

    (BD)(x)=B(x)D(x)=(2.6x2+33x+4043)(2x213x+2428)Substitute the given functions.=2.6x2+33x+4043+(2x3)+13x2428Remove parentheses andchange the sign of each term inthe second set of parentheses.=(2.6x22x2)+(33x+13x)+(40432428)Group like terms.=4.6x2+46x+1615Combine like terms.

    The function

    (BD)(x)=4.6x2+46x+1615

    models the change in U.S. population, in thousands, x years after 2000.

  2. Because 2014 is 14 years after 2000, we substitute 14 for x in the difference function (BD)(x)

    (BD)(x)=4.6x2+46x+1615Use the difference function BD.(BD)(14)=4.6(14)2+46(14)+1615Substitute 14 for x.=4.6(196)+46(14)+1615Evaluate the exponential expression:142=196.=901.6+644+1615Perform the multiplications.=1357.4Add from left to right.

    We see that (BD)(14)=1357.4 The model indicates that there was a population increase of 1357.4 thousand, or approximately 1,357,400 people, in 2014.

  3. The data for 2014 in Figure 1.66 show 3988 thousand births and 2628 thousand deaths.

population change=birthsdeathspopulation change=39882628=1360

The actual population increase was 1360 thousand, or 1,360,000. Our model gave us an increase of 1357.4 thousand. Thus, the model underestimates the actual increase by 13601357.4, or 2.6 thousand people.

Check Point 4

  • Use the birth and death models from Example 4.

    1. Write a function that models the total number of births and deaths in the United States for the years from 2000 through 2016.

    2. Use the function from part (a) to find the total number of births and deaths in the United States in 2000.

    3. Does the result in part (b) overestimate or underestimate the actual number of total births and deaths in 2000 obtained from the data in Figure 1.66? By how much?

Objective 2: Combine functions using the algebra of functions, specifying domains

The Algebra of Functions

  1. Objective 2 Combine functions using the algebra of functions, specifying domains.

Watch Video

We can combine functions using addition, subtraction, multiplication, and division by performing operations with the algebraic expressions that appear on the right side of the equations. For example, the functions f(x)=2x and g(x)=x1 can be combined to form the sum, difference, product, and quotient of f and g. Here’s how it’s done:

The image shows the solutions for the sum, difference, product, and quotient of f and g.
1.7-275 Full Alternative Text

The domain for each of these functions consists of all real numbers that are common to the domains of f and g. Using Df to represent the domain of f and Dg to represent the domain of g, the domain for each function is DfDg. In the case of the quotient function f(x)g(x), we must remember not to divide by 0, so we add the further restriction that g(x)0.

The Algebra of Functions: Sum, Difference, Product, and Quotient of Functions

Let f and g be two functions. The sum f+g, the difference fg, the product fg, and the quotient fg are functions whose domains are the set of all real numbers common to the domains of f and g(DfDg), defined as follows:

1. Sum: (f+g)(x)=f(x)+g(x)
2. Difference: (fg)(x)=f(x)g(x)
3. Product: (fg)(x)=f(x)g(x)
4. Quotient: (fg)(x)=f(x)g(x), provided g(x)0.

Example 2 Combining Functions

Let f(x)=2x1 and g(x)=x2+x2. Find each of the following functions:

  1. (f+g)(x)

  2. (fg)(x)

  3. (fg)(x)

  4. (fg)(x).

Determine the domain for each function.

Solution

  1.  

    (f+g)(x)=f(x)+g(x)This is the definition of the sum f+g.(f+g)(x)=(2x1)+(x2+x2)Substitute the given functions.(f+g)(x)=x2+3x3Remove parentheses and combine like terms.
  2.  

    (fg)(x)=f(x)g(x)This is the definition of the difference fg.(f+g)(x)=(2x1)(x2+x2)Substitute the given functions.(f+g)(x)=2x1x2x+2Remove parentheses and change the sign ofeach term in the second set of parentheses.(f+g)(x)=x2+x+1Combine like terms and arrange terms indescending powers of x.

  3.  

    (fg)(x)=f(x)g(x)This is the definition of theproduct fg.=(2x1)(x2+x+2)Substitute the given functions.=2x(x2+x2)1(x2+x2)Multiply each term in thesecond factor by 2x and 1,respectively.=2x3+2x24xx2x+2Use the distributive property.=2x3+(2x2x2)+(4xx)+2Rearrange terms so that liketerms are adjacent.=2x3+x25x+2Combine like terms.
  4.  

    (fg)(x)=f(x)g(x)This is the definition of the quotient fg.=2x1x2+x2Substitute the given functions. This rational expressioncannot be simplified.

Because the equations for f and g do not involve division or contain even roots, the domain of both f and g is the set of all real numbers. Thus, the domain of f+g, fg, and fg is the set of all real numbers, (, ).

The function fg contains division. We must exclude from its domain values of x that cause the denominator, x2+x2, to be 0. Let’s identify these values.

x2+x2=0Set the denominator of fg equal to 0.(x+2)(x1)=0Factor.x+2=0orx1=0Set each factor equal to 0.x=2x=1Solve the resulting equations.

We must exclude 2 and 1 from the domain of fg.

Domain offg=(, 2)(2, 1)(1, )

Check Point 2

  • Let f(x)=x5 and g(x)=x21. Find each of the following functions:

    1. (f+g)(x)

    2. (fg)(x)

    3. (fg)(x)

    4. (fg)(x).

    Determine the domain for each function.

Example 3 Adding Functions and Determining the Domain

Let f(x)=x+3 and g(x)=x2. Find each of the following:

  1. (f+g)(x)

  2. the domain of f+g.

Solution

  1. (f+g)(x)=f(x)+g(x)=x++x2

  2. The domain of f+g is the set of all real numbers that are common to the domain of f and the domain of g. Thus, we must find the domains of f and g before finding their intersection.

f of x = square root of start expression x + 3 end expression. g of x = square root of start expression x minus 2 end expression.

Now, we can use a number line to determine DfDg, the domain of f+g. Figure 1.68 shows the domain of f, [3, ), in blue and the domain of g, [2, ), in red. Can you see that all real numbers greater than or equal to 2 are common to both domains? This is shown in purple on the number line. Thus, the domain of f+g is [2, ).

Figure 1.68 Finding the domain of the sum f+g

A number line determines the domain of f, the domain of g, and the domain of f + g.

Check Point 3

  • Let f(x)=x3 and g(x)=x+1. Find each of the following:

    1. (f+g)(x)

    2. the domain of f+g.

Example 4 Applying the Algebra of Functions

We opened the section with functions that model the number of births and deaths in the United States for selected years from 2000 through 2016:

B of x = negative 2.6 x squared + 33 x + 4043, labeled, number of births comma B of x comma in thousands comma x years after 2000.
  1. Write a function that models the change in U.S. population, B(x)D(x), for each year from 2000 through 2016.

  2. Use the function from part (a) to find the change in U.S. population in 2014.

  3. Does the result in part (b) overestimate or underestimate the actual population change in 2014 obtained from the data in Figure 1.66? By how much?

Solution

  1. The change in population is the number of births minus the number of deaths. Thus, we will find the difference function, BD.

    (BD)(x)=B(x)D(x)=(2.6x2+33x+4043)(2x213x+2428)Substitute the given functions.=2.6x2+33x+4043+(2x3)+13x2428Remove parentheses andchange the sign of each term inthe second set of parentheses.=(2.6x22x2)+(33x+13x)+(40432428)Group like terms.=4.6x2+46x+1615Combine like terms.

    The function

    (BD)(x)=4.6x2+46x+1615

    models the change in U.S. population, in thousands, x years after 2000.

  2. Because 2014 is 14 years after 2000, we substitute 14 for x in the difference function (BD)(x)

    (BD)(x)=4.6x2+46x+1615Use the difference function BD.(BD)(14)=4.6(14)2+46(14)+1615Substitute 14 for x.=4.6(196)+46(14)+1615Evaluate the exponential expression:142=196.=901.6+644+1615Perform the multiplications.=1357.4Add from left to right.

    We see that (BD)(14)=1357.4 The model indicates that there was a population increase of 1357.4 thousand, or approximately 1,357,400 people, in 2014.

  3. The data for 2014 in Figure 1.66 show 3988 thousand births and 2628 thousand deaths.

population change=birthsdeathspopulation change=39882628=1360

The actual population increase was 1360 thousand, or 1,360,000. Our model gave us an increase of 1357.4 thousand. Thus, the model underestimates the actual increase by 13601357.4, or 2.6 thousand people.

Check Point 4

  • Use the birth and death models from Example 4.

    1. Write a function that models the total number of births and deaths in the United States for the years from 2000 through 2016.

    2. Use the function from part (a) to find the total number of births and deaths in the United States in 2000.

    3. Does the result in part (b) overestimate or underestimate the actual number of total births and deaths in 2000 obtained from the data in Figure 1.66? By how much?

Objective 3: Form composite functions

Composite Functions

  1. Objective 3 Form composite functions.

Watch Video

There is another way of combining two functions. To help understand this new combination, suppose that your local computer store is having a sale. The models that are on sale cost either $300 less than the regular price or 85% of the regular price. If x represents the computer’s regular price, the discounts can be modeled with the following functions:

f of x = x minus 300, x minus 3 is labeled, the computer is on sale for $ 300 less than its regular price. g of x = 0.85 x. 0.85 x is labeled, the computer is on sale for 85 % of its regular price.

At the store, you bargain with the salesperson. Eventually, she makes an offer you can’t refuse. The sale price will be 85% of the regular price followed by a $300 reduction:

0.85 x minus 300. 0.85 x is labeled, 85 % of the regular price. 300 is labeled, followed by a $ 300 reduction.

In terms of the functions f and g, this offer can be obtained by taking the output of g(x)=0.85x, namely, 0.85x, and using it as the input of f:

f of x = x minus 300. Replace x with 0.85 x, the output of g of x = 0.85 x. f of 0.85 x = 0.85 x minus 300.

Because 0.85x is g(x), we can write this last equation as

f(g(x))=0.85x300.

We read this equation as “f of g of x is equal to 0.85x300.” We call f(g(x)) the composition of the function f with g, or a composite function. This composite function is written fg. Thus,

f composed with g of x = f left parenthesis g of x right parenthesis = 0.85 x minus 300.
1.7-283 Full Alternative Text

Like all functions, we can evaluate fg for a specified value of x in the function’s domain. For example, here’s how to find the value of the composite function describing the offer you cannot refuse at 1400:

f composed with g of x = 0.85 x minus 300. Replace x with 1400. Left parenthesis f composed with g right parenthesis, left parenthesis 1400 right parenthesis = 0.85 left parenthesis 1400 right parenthesis minus 300 = 1190 minus 300 = 890.

This means that a computer that regularly sells for $1400 is on sale for $890 subject to both discounts. We can use a partial table of coordinates for each of the discount functions, g and f, to verify this result numerically.

The image shows a table for functions g and f.
1.7-285 Full Alternative Text

Using these tables, we can find (fg)(1400):

Left parenthesis f composed with g right parenthesis, left parenthesis 1400 right parenthesis = f left parenthesis g of 1400 right parenthesis = f of 1190 = 890.
1.7-286 Full Alternative Text

This verifies that a computer that regularly sells for $1400 is on sale for $890 subject to both discounts.

Before you run out to buy a computer, let’s generalize our discussion of the computer’s double discount and define the composition of any two functions.

The Composition of Functions

The composition of the function f with g is denoted by fg and is defined by the equation

(fg)(x)=f(g(x)).

The domain of the composite function fg is the set of all x such that

  1. x is in the domain of g and

  2. g(x) is in the domain of f.

The composition of f with g, fg, is illustrated in Figure 1.69.

  1. Step 1 Input x into g.

  2. Step 2 Input g(x) into f.

Figure 1.69

An illustration depicts a function g yields the output, g of x, with an input, x. Further, it yields f of g of x as output with input, g of x, in a function f.
Figure 1.69 Full Alternative Text

The figure reinforces the fact that the inside function g in f(g(x)) is done first.

Example 5 Forming Composite Functions

Given f(x)=3x4 and g(x)=x22x+6, find each of the following:

  1. (fg)(x)

  2. (gf)(x)

  3. (gf)(1).

Solution

  1. We begin with (fg)(x), the composition of f with g. Because (fg)(x) means f(g(x)), we must replace each occurrence of x in the equation for f with g(x).

    The process to solve the equation f of x = 3 x minus 4.
  2. Next, we find (gf)(x), the composition of g with f. Because (gf)(x) means g(f(x)), we must replace each occurrence of x in the equation for g with f(x).

    The image shows how to solve the equation g of x = x squared minus 2 x + 6.

    =9x2-24x+16-6x+8+6Use (A-B)2=A2-2AB+B2to square 3x-4.=9x2-30x+30Simplify:-24x-6x=-30xand 16+8+6=30.

    Thus, (gf)(x)=9x230x+30. Notice that (fg(x) is not the same function as (gf)(x).

  3. We can use (gf)(x) to find (gf)(1).

    The image shows how to solve the function g composed with f of x = 9 x squared minus 30 x + 30.

    It is also possible to find (gf)(1) without determining (gf)(x).

    The image shows a mathematical expression g composed with f of 1 = g left parenthesis f of 1 right parenthesis = g of negative 1 = 9.

Check Point 5

  • Given f(x)=5x+6 and g(x)=2x2x1, find each of the following:

    1. (fg)(x)

    2. (gf)(x)

    3. (fg)(1).

Objective 3: Form composite functions

Composite Functions

  1. Objective 3 Form composite functions.

Watch Video

There is another way of combining two functions. To help understand this new combination, suppose that your local computer store is having a sale. The models that are on sale cost either $300 less than the regular price or 85% of the regular price. If x represents the computer’s regular price, the discounts can be modeled with the following functions:

f of x = x minus 300, x minus 3 is labeled, the computer is on sale for $ 300 less than its regular price. g of x = 0.85 x. 0.85 x is labeled, the computer is on sale for 85 % of its regular price.

At the store, you bargain with the salesperson. Eventually, she makes an offer you can’t refuse. The sale price will be 85% of the regular price followed by a $300 reduction:

0.85 x minus 300. 0.85 x is labeled, 85 % of the regular price. 300 is labeled, followed by a $ 300 reduction.

In terms of the functions f and g, this offer can be obtained by taking the output of g(x)=0.85x, namely, 0.85x, and using it as the input of f:

f of x = x minus 300. Replace x with 0.85 x, the output of g of x = 0.85 x. f of 0.85 x = 0.85 x minus 300.

Because 0.85x is g(x), we can write this last equation as

f(g(x))=0.85x300.

We read this equation as “f of g of x is equal to 0.85x300.” We call f(g(x)) the composition of the function f with g, or a composite function. This composite function is written fg. Thus,

f composed with g of x = f left parenthesis g of x right parenthesis = 0.85 x minus 300.
1.7-283 Full Alternative Text

Like all functions, we can evaluate fg for a specified value of x in the function’s domain. For example, here’s how to find the value of the composite function describing the offer you cannot refuse at 1400:

f composed with g of x = 0.85 x minus 300. Replace x with 1400. Left parenthesis f composed with g right parenthesis, left parenthesis 1400 right parenthesis = 0.85 left parenthesis 1400 right parenthesis minus 300 = 1190 minus 300 = 890.

This means that a computer that regularly sells for $1400 is on sale for $890 subject to both discounts. We can use a partial table of coordinates for each of the discount functions, g and f, to verify this result numerically.

The image shows a table for functions g and f.
1.7-285 Full Alternative Text

Using these tables, we can find (fg)(1400):

Left parenthesis f composed with g right parenthesis, left parenthesis 1400 right parenthesis = f left parenthesis g of 1400 right parenthesis = f of 1190 = 890.
1.7-286 Full Alternative Text

This verifies that a computer that regularly sells for $1400 is on sale for $890 subject to both discounts.

Before you run out to buy a computer, let’s generalize our discussion of the computer’s double discount and define the composition of any two functions.

The Composition of Functions

The composition of the function f with g is denoted by fg and is defined by the equation

(fg)(x)=f(g(x)).

The domain of the composite function fg is the set of all x such that

  1. x is in the domain of g and

  2. g(x) is in the domain of f.

The composition of f with g, fg, is illustrated in Figure 1.69.

  1. Step 1 Input x into g.

  2. Step 2 Input g(x) into f.

Figure 1.69

An illustration depicts a function g yields the output, g of x, with an input, x. Further, it yields f of g of x as output with input, g of x, in a function f.
Figure 1.69 Full Alternative Text

The figure reinforces the fact that the inside function g in f(g(x)) is done first.

Example 5 Forming Composite Functions

Given f(x)=3x4 and g(x)=x22x+6, find each of the following:

  1. (fg)(x)

  2. (gf)(x)

  3. (gf)(1).

Solution

  1. We begin with (fg)(x), the composition of f with g. Because (fg)(x) means f(g(x)), we must replace each occurrence of x in the equation for f with g(x).

    The process to solve the equation f of x = 3 x minus 4.
  2. Next, we find (gf)(x), the composition of g with f. Because (gf)(x) means g(f(x)), we must replace each occurrence of x in the equation for g with f(x).

    The image shows how to solve the equation g of x = x squared minus 2 x + 6.

    =9x2-24x+16-6x+8+6Use (A-B)2=A2-2AB+B2to square 3x-4.=9x2-30x+30Simplify:-24x-6x=-30xand 16+8+6=30.

    Thus, (gf)(x)=9x230x+30. Notice that (fg(x) is not the same function as (gf)(x).

  3. We can use (gf)(x) to find (gf)(1).

    The image shows how to solve the function g composed with f of x = 9 x squared minus 30 x + 30.

    It is also possible to find (gf)(1) without determining (gf)(x).

    The image shows a mathematical expression g composed with f of 1 = g left parenthesis f of 1 right parenthesis = g of negative 1 = 9.

Check Point 5

  • Given f(x)=5x+6 and g(x)=2x2x1, find each of the following:

    1. (fg)(x)

    2. (gf)(x)

    3. (fg)(1).

Objective 3: Form composite functions

Composite Functions

  1. Objective 3 Form composite functions.

Watch Video

There is another way of combining two functions. To help understand this new combination, suppose that your local computer store is having a sale. The models that are on sale cost either $300 less than the regular price or 85% of the regular price. If x represents the computer’s regular price, the discounts can be modeled with the following functions:

f of x = x minus 300, x minus 3 is labeled, the computer is on sale for $ 300 less than its regular price. g of x = 0.85 x. 0.85 x is labeled, the computer is on sale for 85 % of its regular price.

At the store, you bargain with the salesperson. Eventually, she makes an offer you can’t refuse. The sale price will be 85% of the regular price followed by a $300 reduction:

0.85 x minus 300. 0.85 x is labeled, 85 % of the regular price. 300 is labeled, followed by a $ 300 reduction.

In terms of the functions f and g, this offer can be obtained by taking the output of g(x)=0.85x, namely, 0.85x, and using it as the input of f:

f of x = x minus 300. Replace x with 0.85 x, the output of g of x = 0.85 x. f of 0.85 x = 0.85 x minus 300.

Because 0.85x is g(x), we can write this last equation as

f(g(x))=0.85x300.

We read this equation as “f of g of x is equal to 0.85x300.” We call f(g(x)) the composition of the function f with g, or a composite function. This composite function is written fg. Thus,

f composed with g of x = f left parenthesis g of x right parenthesis = 0.85 x minus 300.
1.7-283 Full Alternative Text

Like all functions, we can evaluate fg for a specified value of x in the function’s domain. For example, here’s how to find the value of the composite function describing the offer you cannot refuse at 1400:

f composed with g of x = 0.85 x minus 300. Replace x with 1400. Left parenthesis f composed with g right parenthesis, left parenthesis 1400 right parenthesis = 0.85 left parenthesis 1400 right parenthesis minus 300 = 1190 minus 300 = 890.

This means that a computer that regularly sells for $1400 is on sale for $890 subject to both discounts. We can use a partial table of coordinates for each of the discount functions, g and f, to verify this result numerically.

The image shows a table for functions g and f.
1.7-285 Full Alternative Text

Using these tables, we can find (fg)(1400):

Left parenthesis f composed with g right parenthesis, left parenthesis 1400 right parenthesis = f left parenthesis g of 1400 right parenthesis = f of 1190 = 890.
1.7-286 Full Alternative Text

This verifies that a computer that regularly sells for $1400 is on sale for $890 subject to both discounts.

Before you run out to buy a computer, let’s generalize our discussion of the computer’s double discount and define the composition of any two functions.

The Composition of Functions

The composition of the function f with g is denoted by fg and is defined by the equation

(fg)(x)=f(g(x)).

The domain of the composite function fg is the set of all x such that

  1. x is in the domain of g and

  2. g(x) is in the domain of f.

The composition of f with g, fg, is illustrated in Figure 1.69.

  1. Step 1 Input x into g.

  2. Step 2 Input g(x) into f.

Figure 1.69

An illustration depicts a function g yields the output, g of x, with an input, x. Further, it yields f of g of x as output with input, g of x, in a function f.
Figure 1.69 Full Alternative Text

The figure reinforces the fact that the inside function g in f(g(x)) is done first.

Example 5 Forming Composite Functions

Given f(x)=3x4 and g(x)=x22x+6, find each of the following:

  1. (fg)(x)

  2. (gf)(x)

  3. (gf)(1).

Solution

  1. We begin with (fg)(x), the composition of f with g. Because (fg)(x) means f(g(x)), we must replace each occurrence of x in the equation for f with g(x).

    The process to solve the equation f of x = 3 x minus 4.
  2. Next, we find (gf)(x), the composition of g with f. Because (gf)(x) means g(f(x)), we must replace each occurrence of x in the equation for g with f(x).

    The image shows how to solve the equation g of x = x squared minus 2 x + 6.

    =9x2-24x+16-6x+8+6Use (A-B)2=A2-2AB+B2to square 3x-4.=9x2-30x+30Simplify:-24x-6x=-30xand 16+8+6=30.

    Thus, (gf)(x)=9x230x+30. Notice that (fg(x) is not the same function as (gf)(x).

  3. We can use (gf)(x) to find (gf)(1).

    The image shows how to solve the function g composed with f of x = 9 x squared minus 30 x + 30.

    It is also possible to find (gf)(1) without determining (gf)(x).

    The image shows a mathematical expression g composed with f of 1 = g left parenthesis f of 1 right parenthesis = g of negative 1 = 9.

Check Point 5

  • Given f(x)=5x+6 and g(x)=2x2x1, find each of the following:

    1. (fg)(x)

    2. (gf)(x)

    3. (fg)(1).

Objective 4: Determine domains for composite functions

  1. Objective 4 Determine domains for composite functions.

Watch Video

We need to be careful in determining the domain for a composite function.

Excluding Values from the Domain of (f°g)(x)=f(g(x))

The following values must be excluded from the input x:

Example 6 Forming a Composite Function and Finding Its Domain

Given f(x)=2x1 and g(x)=3x, find each of the following:

  1. (fg)(x)

  2. the domain of fg.

Solution

  1. Because (fg)(x) means f(g(x)), we must replace x in f(x)=2x1 with g(x).

    The image shows the process to find the composite function f composed with g of x.

    Thus, (fg)(x)=2x3x.

  2. We determine values to exclude from the domain of (fg)(x) in two steps.

    Rules for Excluding Numbers from the Domain of (f ° g)(x)=f(g(x)) Applying the Rules to f(x)=2x1 and g(x)=3x
    If x is not in the domain of g, it must not be in the domain of fg. Because g(x)=3x, 0 is not in the domain of g. Thus, 0 must be excluded from the domain of fg.
    Any x for which g(x) is not in the domain of f must not be in the domain of fg.

    Because f(g(x))=2g(x)1, we must exclude from the domain of fg any x for which g(x)=1.

    .3x=1 Set  g(x)equal to 1. 3=x Multiply both sides by  x.

    3 must be excluded from the domain of fg.

    We see that 0 and 3 must be excluded from the domain of fg. The domain of fg is

    (, 0)(0, 3)(3, ).

Check Point 6

  • Given f(x)=4x+2 and g(x)=1x, find each of the following:

    1. (fg)(x)

    2. the domain of fg.

Objective 5: Write functions as compositions

Decomposing Functions

  1. Objective 5 Write functions as compositions.

Watch Video

When you form a composite function, you “compose” two functions to form a new function. It is also possible to reverse this process. That is, you can “decompose” a given function and express it as a composition of two functions. Although there is more than one way to do this, there is often a “natural” selection that comes to mind first. For example, consider the function h defined by

h(x)=(3x24x+1)5.

The function h takes 3x24x+1 and raises it to the power 5. A natural way to write h as a composition of two functions is to raise the function g (x)=3x24x+1 to the power 5. Thus, if we let

g....f (x)=x5 and g(x)=3x24x+1,then  (fg)(x)=f(g(x))=f(3x24x+1)=(3x24x+1)5.

Example 7 Writing a Function as a Composition

Express h(x) as a composition of two functions:

h(x)=x2+13.

Solution

The function h takes x2+1 and takes its cube root. A natural way to write h as a composition of two functions is to take the cube root of the function g(x)=x2+1. Thus, we let

f(x)=x3 and g(x)=x2+1.

We can check this composition by finding (fg)(x). This should give the original function, namely, h(x)=x2+13.

(fg)(x)=f(g(x))=f(x2+1)=x2+13=h(x)

Check Point 7

  • Express h(x) as a composition of two functions:

    h(x)=x2+5.
1.7: Exercise Set

1.7 Exercise Set

Practice Exercises

In Exercises 130, find the domain of each function.

  1. 1. f(x)=3(x4)

  2. 2. f(x)=2(x+5)

  3. 3. g(x)=3x4

  4. 4. g(x)=2x+5

  5. 5. f(x)=x22x15

  6. 6. f(x)=x2+x12

  7. 7. g(x)=3x22x15

  8. 8. g(x)=2x2+x12

  9. 9. f(x)=1x+7+3x9

  10. 10. f(x)=1x+8+3x10

  11. 11. g(x)=1x2+11x21

  12. 12. g(x)=1x2+41x24

  13. 13. h(x)=43x1

  14. 14. h(x)=54x1

  15. 15. f(x)=14x12

  16. 16. f(x)=14x23

  17. 17. f(x)=x3

  18. 18. f(x)=x+2

  19. 19. g(x)=1x3

  20. 20. g(x)=1x+2

  21. 21. g(x)=5x+35

  22. 22. g(x)=7x70

  23. 23. f(x)=242x

  24. 24. f(x)=846x

  25. 25. h(x)=x2+x+3

  26. 26. h(x)=x3+x+4

  27. 27. g(x)=x2x5

  28. 28. g(x)=x3x6

  29. 29. f(x)=2x+7x35x24x+20

  30. 30. f(x)=7x+2x32x29x+18

In Exercises 3150, find f+g, fg, fg, and fg. Determine the domain for each function.

  1. 31. f(x)=2x+3, g(x)=x1

  2. 32. f(x)=3x4, g(x)=x+2

  3. 33. f(x)=x5, g(x)=3x2

  4. 34. f(x)=x6, g(x)=5x2

  5. 35. f(x)=2x2x3, g(x)=x+1

  6. 36. f(x)=6x2x1, g(x)=x1

  7. 37. f(x)=3x2, g(x)=x2+2x15

  8. 38. f(x)=5x2, g(x)=x2+4x12

  9. 39. f(x)=x, g(x)=x4

  10. 40. f(x)=x, g(x)=x5

  11. 41. f(x)=2+1x, g(x)=1x

  12. 42. f(x)=61x, g(x)=1x

  13. 43. f(x)=5x+1x29, g(x)=4x2x29

  14. 44. f(x)=3x+1x225, g(x)=2x4x225

  15. 45. f(x)=8xx2, g(x)=6x+3

  16. 46. f(x)=9xx4, g(x)=7x+8

  17. 47. f(x)=x+4, g(x)=x1

  18. 48. f(x)=x+6, g(x)=x3

  19. 49. f(x)=x2, g(x)=2x

  20. 50. f(x)=x5, g(x)=5x

In Exercises 5166, find

  1. (fg)(x)

  2. (gf)(x)

  3. (fg)(2)

  4. (gf)(2).

  1. 51. f(x)=2x, g(x)=x+7

  2. 52. f(x)=3x, g(x)=x5

  3. 53. f(x)=x+4, g(x)=2x+1

  4. 54. f(x)=5x+2, g(x)=3x4

  5. 55. f(x)=4x3, g(x)=5x22

  6. 56. f(x)=7x+1, g(x)=2x29

  7. 57. f(x)=x2+2, g(x)=x22

  8. 58. f(x)=x2+1, g(x)=x23

  9. 59. f(x)=4x, g(x)=2x2+x+5

  10. 60. f(x)=5x2, g(x)=x2+4x1

  11. 61. f(x)=x, g(x)=x1

  12. 62. f(x)=x, g(x)=x+2

  13. 63. f(x)=2x3, g(x)=x+32

  14. 64. f(x)=6x3, g(x)=x+36

  15. 65. f(x)=1x, g(x)=1x

  16. 66. f(x)=2x, g(x)=2x

In Exercises 6774, find

  1. (fg)(x)

  2. the domain of fg.

  1. 67. f(x)=2x+3, g(x)=1x

  2. 68. f(x)=5x+4, g(x)=1x

  3. 69. f(x)=xx+1, g(x)=4x

  4. 70. f(x)=xx+5, g(x)=6x

  5. 71. f(x)=x, g(x)=x2

  6. 72. f(x)=x, g(x)=x3

  7. 73. f(x)=x2+4, g(x)=1x

  8. 74. f(x)=x2+1, g(x)=2x

In Exercises 7582, express the given function h as a composition of two functions f and g so that h(x)=(fg)(x).

  1. 75. h(x)=(3x1)4

  2. 76. h(x)=(2x5)3

  3. 77. h(x)=x293

  4. 78. h(x)=5x2+3

  5. 79. h(x)=|2x5|

  6. 80. h(x)=|3x4|

  7. 81. h(x)=12x3

  8. 82. h(x)=14x+5

Practice PLUS

Use the graphs of f and g to solve Exercises 8390.

The image shows the graph that plots two fluctuating lines, y = f of x and y = g of x.
  1. 83. Find (f+g)(3).

  2. 84. Find (gf)(2).

  3. 85. Find (fg)(2).

  4. 86. Find (gf)(3).

  5. 87. Find the domain of f+g.

  6. 88. Find the domain of fg.

  7. 89. Graph f+g.

  8. 90. Graph fg.

In Exercises 9194, use the graphs of f and g to evaluate each composite function.

The image shows a graph that plots two lines.
  1. 91. (fg)(1)

  2. 92. (fg)(1)

  3. 93. (gf)(0)

  4. 94. (gf)(1)

In Exercises 9596, find all values of x satisfying the given conditions.

  1. 95. f(x)=2x5, g(x)=x23x+8, and (fg)(x)=7.

  2. 96. f(x)=12x, g(x)=3x2+x1, and (fg)(x)=5.

Application Exercises

The bar graph shows U.S. population projections, by age, in millions, for five selected years.

The image shows a bar graph titled, United States population projections by age for under 45 and 4 and older. The graph plots population in millions versus year.

Source: U.S. Census Bureau

Here are two functions that model the data in the graph at the bottom of the previous column:

Two expressions.

Use the functions to solve Exercises 9798.

  1. 97.

    1. Write a function d that models the difference between the projected population under 45 and the projected population 45 and older for the years shown in the bar graph.

    2. Use the function from part (a) to find how many more people under 45 than 45 and older there are projected to be in 2060.

    3. Does the result in part (b) overestimate, underestimate, or give the actual difference between the under-45 and 45-and-older populations in 2060 shown by the bar graph?

  2. 98.

    1. Write a function r that models the ratio of the projected population 45 and older to the projected population under 45 for the years shown in the bar graph.

    2. Use the function from part (a) to find the ratio of the projected population 45 and older to the projected population under 45, correct to two decimal places, for 2040.

    3. Find the ratio of the projected population 45 and older to the projected population under 45, correct to two decimal places, shown by the bar graph. How does the rounded ratio in part (b) compare with this ratio?

  3. 99. A company that sells protective tablet cases has yearly fixed costs of $600,000. It costs the company $45 to produce each case. Each case will sell for $65. The company’s costs and revenue are modeled by the following functions, where x represents the number of tablet cases produced and sold:

    C (x)=600,000+45xThis function models thecompanys costs.R(x)=65x.This function models thecompanys revenue.

    Find and interpret (RC)(20,000), (RC)(30,000), and (RC)(40,000).

  4. 100. A department store has two locations in a city. From 2016 through 2020, the profits for each of the store’s two branches are modeled by the functions f(x)=0.44x+13.62 and g(x)=0.51x+11.14. In each model, x represents the number of years after 2016, and f and g represent the profit, in millions of dollars.

    1. What is the slope of f? Describe what this means.

    2. What is the slope of g? Describe what this means.

    3. Find f+g. What is the slope of this function? What does this mean?

  5. 101. The regular price of a laptop is x dollars. Let f(x)=x400 and g(x)=0.75x.

    1. Describe what the functions f and g model in terms of the price of the laptop.

    2. Find (fg)(x) and describe what this models in terms of the price of the laptop.

    3. Repeat part (b) for (gf)(x).

    4. Which composite function models the greater discount on the laptop, fg or gf? Explain.

  6. 102. The regular price of a pair of jeans is x dollars. Let f(x)=x5 and g(x)=0.6x.

    1. Describe what functions f and g model in terms of the price of the jeans.

    2. Find (fg)(x) and describe what this models in terms of the price of the jeans.

    3. Repeat part (b) for (gf)(x).

    4. Which composite function models the greater discount on the jeans, fg or gf? Explain.

Explaining the Concepts

  1. 103. If a function is defined by an equation, explain how to find its domain.

  2. 104. If equations for f and g are given, explain how to find fg.

  3. 105. If equations for two functions are given, explain how to obtain the quotient function and its domain.

  4. 106. Describe a procedure for finding (fg)(x). What is the name of this function?

  5. 107. Describe the values of x that must be excluded from the domain of (fg)(x).

Technology Exercises

  1. 108. Graph y1=x22x, y2=x, and y3=y1÷y2 in the same [10, 10, 1] by [10, 10, 1] viewing rectangle. Then use the TRACE feature to trace along y3. What happens at x=0? Explain why this occurs.

  2. 109. Graph y1=2x, y2=x, and y3=2y2 in the same [4, 4, 1] by [0, 2, 1] viewing rectangle. If y1 represents f and y2 represents g, use the graph of y3 to find the domain of fg. Then verify your observation algebraically.

Critical Thinking Exercises

Make Sense? In Exercises 110113, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 110. I used a function to model data from 1990 through 2020. The independent variable in my model represented the number of years after 1990, so the function’s domain was {x|x=0, 1, 2, 3,, 30}.

  2. 111. I have two functions. Function f models total world population x years after 2000 and function g models population of the world’s more-developed regions x years after 2000. I can use fg to determine the population of the world’s less-developed regions for the years in both function’s domains.

  3. 112. I must have made a mistake in finding the composite functions fg and gf, because I notice that fg is not the same function as gf.

  4. 113. This diagram illustrates that f(g(x))=x2+4.

    An illustration depicts a function f of x, x squared yields the first output, x squared, with a first input, x and function g of x, x + 4 results the second output x squared + 4 with a double input x squared.

In Exercises 114117, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 114. If f(x)=x24 and g(x)=x24, then (fg)(x)=x2 and (fg)(5)=25.

  2. 115. There can never be two functions f and g, where fg, for which (fg)(x)=(gf)(x).

  3. 116. If f(7)=5 and g(4)=7, then (fg)(4)=35.

  4. 117. If f(x)=x and g(x)=2x1, then (fg)(5)=g(2).

  5. 118. Prove that if f and g are even functions, then fg is also an even function.

  6. 119. Define two functions f and g so that fg=gf.

Retaining the Concepts

  1. 120. Solve and check: x15x+32=1x4.

    (Section P.7, Example 2)

  2. 121. In July 2020, the toll for the Golden Gate Bridge was $8.40 for drivers making one-time payments for each crossing. Drivers who pay a one-time fee of $20.00 to purchase a FasTrak toll tag pay $7.70 for each crossing. How many times must a driver cross the Golden Gate Bridge for the cost of these two options to be the same? Round to the nearest whole number. Find the total cost of each option for the rounded number of crossings.

    (Section P.8, Example 3)

  3. 122. Solve for y: Ax+By=Cy+D.

    (Section P.7, Example 5)

Preview Exercises

Exercises 123125 will help you prepare for the material covered in the next section.

  1. 123. Consider the function defined by

    {(2, 4), (1, 1), (1, 1), (2, 4)}.

    Reverse the components of each ordered pair and write the resulting relation. Is this relation a function?

  2. 124. Solve for y:x=5y+4.

  3. 125. Solve for y:x=y21, y0.

1.7: Exercise Set

1.7 Exercise Set

Practice Exercises

In Exercises 130, find the domain of each function.

  1. 1. f(x)=3(x4)

  2. 2. f(x)=2(x+5)

  3. 3. g(x)=3x4

  4. 4. g(x)=2x+5

  5. 5. f(x)=x22x15

  6. 6. f(x)=x2+x12

  7. 7. g(x)=3x22x15

  8. 8. g(x)=2x2+x12

  9. 9. f(x)=1x+7+3x9

  10. 10. f(x)=1x+8+3x10

  11. 11. g(x)=1x2+11x21

  12. 12. g(x)=1x2+41x24

  13. 13. h(x)=43x1

  14. 14. h(x)=54x1

  15. 15. f(x)=14x12

  16. 16. f(x)=14x23

  17. 17. f(x)=x3

  18. 18. f(x)=x+2

  19. 19. g(x)=1x3

  20. 20. g(x)=1x+2

  21. 21. g(x)=5x+35

  22. 22. g(x)=7x70

  23. 23. f(x)=242x

  24. 24. f(x)=846x

  25. 25. h(x)=x2+x+3

  26. 26. h(x)=x3+x+4

  27. 27. g(x)=x2x5

  28. 28. g(x)=x3x6

  29. 29. f(x)=2x+7x35x24x+20

  30. 30. f(x)=7x+2x32x29x+18

In Exercises 3150, find f+g, fg, fg, and fg. Determine the domain for each function.

  1. 31. f(x)=2x+3, g(x)=x1

  2. 32. f(x)=3x4, g(x)=x+2

  3. 33. f(x)=x5, g(x)=3x2

  4. 34. f(x)=x6, g(x)=5x2

  5. 35. f(x)=2x2x3, g(x)=x+1

  6. 36. f(x)=6x2x1, g(x)=x1

  7. 37. f(x)=3x2, g(x)=x2+2x15

  8. 38. f(x)=5x2, g(x)=x2+4x12

  9. 39. f(x)=x, g(x)=x4

  10. 40. f(x)=x, g(x)=x5

  11. 41. f(x)=2+1x, g(x)=1x

  12. 42. f(x)=61x, g(x)=1x

  13. 43. f(x)=5x+1x29, g(x)=4x2x29

  14. 44. f(x)=3x+1x225, g(x)=2x4x225

  15. 45. f(x)=8xx2, g(x)=6x+3

  16. 46. f(x)=9xx4, g(x)=7x+8

  17. 47. f(x)=x+4, g(x)=x1

  18. 48. f(x)=x+6, g(x)=x3

  19. 49. f(x)=x2, g(x)=2x

  20. 50. f(x)=x5, g(x)=5x

In Exercises 5166, find

  1. (fg)(x)

  2. (gf)(x)

  3. (fg)(2)

  4. (gf)(2).

  1. 51. f(x)=2x, g(x)=x+7

  2. 52. f(x)=3x, g(x)=x5

  3. 53. f(x)=x+4, g(x)=2x+1

  4. 54. f(x)=5x+2, g(x)=3x4

  5. 55. f(x)=4x3, g(x)=5x22

  6. 56. f(x)=7x+1, g(x)=2x29

  7. 57. f(x)=x2+2, g(x)=x22

  8. 58. f(x)=x2+1, g(x)=x23

  9. 59. f(x)=4x, g(x)=2x2+x+5

  10. 60. f(x)=5x2, g(x)=x2+4x1

  11. 61. f(x)=x, g(x)=x1

  12. 62. f(x)=x, g(x)=x+2

  13. 63. f(x)=2x3, g(x)=x+32

  14. 64. f(x)=6x3, g(x)=x+36

  15. 65. f(x)=1x, g(x)=1x

  16. 66. f(x)=2x, g(x)=2x

In Exercises 6774, find

  1. (fg)(x)

  2. the domain of fg.

  1. 67. f(x)=2x+3, g(x)=1x

  2. 68. f(x)=5x+4, g(x)=1x

  3. 69. f(x)=xx+1, g(x)=4x

  4. 70. f(x)=xx+5, g(x)=6x

  5. 71. f(x)=x, g(x)=x2

  6. 72. f(x)=x, g(x)=x3

  7. 73. f(x)=x2+4, g(x)=1x

  8. 74. f(x)=x2+1, g(x)=2x

In Exercises 7582, express the given function h as a composition of two functions f and g so that h(x)=(fg)(x).

  1. 75. h(x)=(3x1)4

  2. 76. h(x)=(2x5)3

  3. 77. h(x)=x293

  4. 78. h(x)=5x2+3

  5. 79. h(x)=|2x5|

  6. 80. h(x)=|3x4|

  7. 81. h(x)=12x3

  8. 82. h(x)=14x+5

Practice PLUS

Use the graphs of f and g to solve Exercises 8390.

The image shows the graph that plots two fluctuating lines, y = f of x and y = g of x.
  1. 83. Find (f+g)(3).

  2. 84. Find (gf)(2).

  3. 85. Find (fg)(2).

  4. 86. Find (gf)(3).

  5. 87. Find the domain of f+g.

  6. 88. Find the domain of fg.

  7. 89. Graph f+g.

  8. 90. Graph fg.

In Exercises 9194, use the graphs of f and g to evaluate each composite function.

The image shows a graph that plots two lines.
  1. 91. (fg)(1)

  2. 92. (fg)(1)

  3. 93. (gf)(0)

  4. 94. (gf)(1)

In Exercises 9596, find all values of x satisfying the given conditions.

  1. 95. f(x)=2x5, g(x)=x23x+8, and (fg)(x)=7.

  2. 96. f(x)=12x, g(x)=3x2+x1, and (fg)(x)=5.

Application Exercises

The bar graph shows U.S. population projections, by age, in millions, for five selected years.

The image shows a bar graph titled, United States population projections by age for under 45 and 4 and older. The graph plots population in millions versus year.

Source: U.S. Census Bureau

Here are two functions that model the data in the graph at the bottom of the previous column:

Two expressions.

Use the functions to solve Exercises 9798.

  1. 97.

    1. Write a function d that models the difference between the projected population under 45 and the projected population 45 and older for the years shown in the bar graph.

    2. Use the function from part (a) to find how many more people under 45 than 45 and older there are projected to be in 2060.

    3. Does the result in part (b) overestimate, underestimate, or give the actual difference between the under-45 and 45-and-older populations in 2060 shown by the bar graph?

  2. 98.

    1. Write a function r that models the ratio of the projected population 45 and older to the projected population under 45 for the years shown in the bar graph.

    2. Use the function from part (a) to find the ratio of the projected population 45 and older to the projected population under 45, correct to two decimal places, for 2040.

    3. Find the ratio of the projected population 45 and older to the projected population under 45, correct to two decimal places, shown by the bar graph. How does the rounded ratio in part (b) compare with this ratio?

  3. 99. A company that sells protective tablet cases has yearly fixed costs of $600,000. It costs the company $45 to produce each case. Each case will sell for $65. The company’s costs and revenue are modeled by the following functions, where x represents the number of tablet cases produced and sold:

    C (x)=600,000+45xThis function models thecompanys costs.R(x)=65x.This function models thecompanys revenue.

    Find and interpret (RC)(20,000), (RC)(30,000), and (RC)(40,000).

  4. 100. A department store has two locations in a city. From 2016 through 2020, the profits for each of the store’s two branches are modeled by the functions f(x)=0.44x+13.62 and g(x)=0.51x+11.14. In each model, x represents the number of years after 2016, and f and g represent the profit, in millions of dollars.

    1. What is the slope of f? Describe what this means.

    2. What is the slope of g? Describe what this means.

    3. Find f+g. What is the slope of this function? What does this mean?

  5. 101. The regular price of a laptop is x dollars. Let f(x)=x400 and g(x)=0.75x.

    1. Describe what the functions f and g model in terms of the price of the laptop.

    2. Find (fg)(x) and describe what this models in terms of the price of the laptop.

    3. Repeat part (b) for (gf)(x).

    4. Which composite function models the greater discount on the laptop, fg or gf? Explain.

  6. 102. The regular price of a pair of jeans is x dollars. Let f(x)=x5 and g(x)=0.6x.

    1. Describe what functions f and g model in terms of the price of the jeans.

    2. Find (fg)(x) and describe what this models in terms of the price of the jeans.

    3. Repeat part (b) for (gf)(x).

    4. Which composite function models the greater discount on the jeans, fg or gf? Explain.

Explaining the Concepts

  1. 103. If a function is defined by an equation, explain how to find its domain.

  2. 104. If equations for f and g are given, explain how to find fg.

  3. 105. If equations for two functions are given, explain how to obtain the quotient function and its domain.

  4. 106. Describe a procedure for finding (fg)(x). What is the name of this function?

  5. 107. Describe the values of x that must be excluded from the domain of (fg)(x).

Technology Exercises

  1. 108. Graph y1=x22x, y2=x, and y3=y1÷y2 in the same [10, 10, 1] by [10, 10, 1] viewing rectangle. Then use the TRACE feature to trace along y3. What happens at x=0? Explain why this occurs.

  2. 109. Graph y1=2x, y2=x, and y3=2y2 in the same [4, 4, 1] by [0, 2, 1] viewing rectangle. If y1 represents f and y2 represents g, use the graph of y3 to find the domain of fg. Then verify your observation algebraically.

Critical Thinking Exercises

Make Sense? In Exercises 110113, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 110. I used a function to model data from 1990 through 2020. The independent variable in my model represented the number of years after 1990, so the function’s domain was {x|x=0, 1, 2, 3,, 30}.

  2. 111. I have two functions. Function f models total world population x years after 2000 and function g models population of the world’s more-developed regions x years after 2000. I can use fg to determine the population of the world’s less-developed regions for the years in both function’s domains.

  3. 112. I must have made a mistake in finding the composite functions fg and gf, because I notice that fg is not the same function as gf.

  4. 113. This diagram illustrates that f(g(x))=x2+4.

    An illustration depicts a function f of x, x squared yields the first output, x squared, with a first input, x and function g of x, x + 4 results the second output x squared + 4 with a double input x squared.

In Exercises 114117, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 114. If f(x)=x24 and g(x)=x24, then (fg)(x)=x2 and (fg)(5)=25.

  2. 115. There can never be two functions f and g, where fg, for which (fg)(x)=(gf)(x).

  3. 116. If f(7)=5 and g(4)=7, then (fg)(4)=35.

  4. 117. If f(x)=x and g(x)=2x1, then (fg)(5)=g(2).

  5. 118. Prove that if f and g are even functions, then fg is also an even function.

  6. 119. Define two functions f and g so that fg=gf.

Retaining the Concepts

  1. 120. Solve and check: x15x+32=1x4.

    (Section P.7, Example 2)

  2. 121. In July 2020, the toll for the Golden Gate Bridge was $8.40 for drivers making one-time payments for each crossing. Drivers who pay a one-time fee of $20.00 to purchase a FasTrak toll tag pay $7.70 for each crossing. How many times must a driver cross the Golden Gate Bridge for the cost of these two options to be the same? Round to the nearest whole number. Find the total cost of each option for the rounded number of crossings.

    (Section P.8, Example 3)

  3. 122. Solve for y: Ax+By=Cy+D.

    (Section P.7, Example 5)

Preview Exercises

Exercises 123125 will help you prepare for the material covered in the next section.

  1. 123. Consider the function defined by

    {(2, 4), (1, 1), (1, 1), (2, 4)}.

    Reverse the components of each ordered pair and write the resulting relation. Is this relation a function?

  2. 124. Solve for y:x=5y+4.

  3. 125. Solve for y:x=y21, y0.

1.7: Exercise Set

1.7 Exercise Set

Practice Exercises

In Exercises 130, find the domain of each function.

  1. 1. f(x)=3(x4)

  2. 2. f(x)=2(x+5)

  3. 3. g(x)=3x4

  4. 4. g(x)=2x+5

  5. 5. f(x)=x22x15

  6. 6. f(x)=x2+x12

  7. 7. g(x)=3x22x15

  8. 8. g(x)=2x2+x12

  9. 9. f(x)=1x+7+3x9

  10. 10. f(x)=1x+8+3x10

  11. 11. g(x)=1x2+11x21

  12. 12. g(x)=1x2+41x24

  13. 13. h(x)=43x1

  14. 14. h(x)=54x1

  15. 15. f(x)=14x12

  16. 16. f(x)=14x23

  17. 17. f(x)=x3

  18. 18. f(x)=x+2

  19. 19. g(x)=1x3

  20. 20. g(x)=1x+2

  21. 21. g(x)=5x+35

  22. 22. g(x)=7x70

  23. 23. f(x)=242x

  24. 24. f(x)=846x

  25. 25. h(x)=x2+x+3

  26. 26. h(x)=x3+x+4

  27. 27. g(x)=x2x5

  28. 28. g(x)=x3x6

  29. 29. f(x)=2x+7x35x24x+20

  30. 30. f(x)=7x+2x32x29x+18

In Exercises 3150, find f+g, fg, fg, and fg. Determine the domain for each function.

  1. 31. f(x)=2x+3, g(x)=x1

  2. 32. f(x)=3x4, g(x)=x+2

  3. 33. f(x)=x5, g(x)=3x2

  4. 34. f(x)=x6, g(x)=5x2

  5. 35. f(x)=2x2x3, g(x)=x+1

  6. 36. f(x)=6x2x1, g(x)=x1

  7. 37. f(x)=3x2, g(x)=x2+2x15

  8. 38. f(x)=5x2, g(x)=x2+4x12

  9. 39. f(x)=x, g(x)=x4

  10. 40. f(x)=x, g(x)=x5

  11. 41. f(x)=2+1x, g(x)=1x

  12. 42. f(x)=61x, g(x)=1x

  13. 43. f(x)=5x+1x29, g(x)=4x2x29

  14. 44. f(x)=3x+1x225, g(x)=2x4x225

  15. 45. f(x)=8xx2, g(x)=6x+3

  16. 46. f(x)=9xx4, g(x)=7x+8

  17. 47. f(x)=x+4, g(x)=x1

  18. 48. f(x)=x+6, g(x)=x3

  19. 49. f(x)=x2, g(x)=2x

  20. 50. f(x)=x5, g(x)=5x

In Exercises 5166, find

  1. (fg)(x)

  2. (gf)(x)

  3. (fg)(2)

  4. (gf)(2).

  1. 51. f(x)=2x, g(x)=x+7

  2. 52. f(x)=3x, g(x)=x5

  3. 53. f(x)=x+4, g(x)=2x+1

  4. 54. f(x)=5x+2, g(x)=3x4

  5. 55. f(x)=4x3, g(x)=5x22

  6. 56. f(x)=7x+1, g(x)=2x29

  7. 57. f(x)=x2+2, g(x)=x22

  8. 58. f(x)=x2+1, g(x)=x23

  9. 59. f(x)=4x, g(x)=2x2+x+5

  10. 60. f(x)=5x2, g(x)=x2+4x1

  11. 61. f(x)=x, g(x)=x1

  12. 62. f(x)=x, g(x)=x+2

  13. 63. f(x)=2x3, g(x)=x+32

  14. 64. f(x)=6x3, g(x)=x+36

  15. 65. f(x)=1x, g(x)=1x

  16. 66. f(x)=2x, g(x)=2x

In Exercises 6774, find

  1. (fg)(x)

  2. the domain of fg.

  1. 67. f(x)=2x+3, g(x)=1x

  2. 68. f(x)=5x+4, g(x)=1x

  3. 69. f(x)=xx+1, g(x)=4x

  4. 70. f(x)=xx+5, g(x)=6x

  5. 71. f(x)=x, g(x)=x2

  6. 72. f(x)=x, g(x)=x3

  7. 73. f(x)=x2+4, g(x)=1x

  8. 74. f(x)=x2+1, g(x)=2x

In Exercises 7582, express the given function h as a composition of two functions f and g so that h(x)=(fg)(x).

  1. 75. h(x)=(3x1)4

  2. 76. h(x)=(2x5)3

  3. 77. h(x)=x293

  4. 78. h(x)=5x2+3

  5. 79. h(x)=|2x5|

  6. 80. h(x)=|3x4|

  7. 81. h(x)=12x3

  8. 82. h(x)=14x+5

Practice PLUS

Use the graphs of f and g to solve Exercises 8390.

The image shows the graph that plots two fluctuating lines, y = f of x and y = g of x.
  1. 83. Find (f+g)(3).

  2. 84. Find (gf)(2).

  3. 85. Find (fg)(2).

  4. 86. Find (gf)(3).

  5. 87. Find the domain of f+g.

  6. 88. Find the domain of fg.

  7. 89. Graph f+g.

  8. 90. Graph fg.

In Exercises 9194, use the graphs of f and g to evaluate each composite function.

The image shows a graph that plots two lines.
  1. 91. (fg)(1)

  2. 92. (fg)(1)

  3. 93. (gf)(0)

  4. 94. (gf)(1)

In Exercises 9596, find all values of x satisfying the given conditions.

  1. 95. f(x)=2x5, g(x)=x23x+8, and (fg)(x)=7.

  2. 96. f(x)=12x, g(x)=3x2+x1, and (fg)(x)=5.

Application Exercises

The bar graph shows U.S. population projections, by age, in millions, for five selected years.

The image shows a bar graph titled, United States population projections by age for under 45 and 4 and older. The graph plots population in millions versus year.

Source: U.S. Census Bureau

Here are two functions that model the data in the graph at the bottom of the previous column:

Two expressions.

Use the functions to solve Exercises 9798.

  1. 97.

    1. Write a function d that models the difference between the projected population under 45 and the projected population 45 and older for the years shown in the bar graph.

    2. Use the function from part (a) to find how many more people under 45 than 45 and older there are projected to be in 2060.

    3. Does the result in part (b) overestimate, underestimate, or give the actual difference between the under-45 and 45-and-older populations in 2060 shown by the bar graph?

  2. 98.

    1. Write a function r that models the ratio of the projected population 45 and older to the projected population under 45 for the years shown in the bar graph.

    2. Use the function from part (a) to find the ratio of the projected population 45 and older to the projected population under 45, correct to two decimal places, for 2040.

    3. Find the ratio of the projected population 45 and older to the projected population under 45, correct to two decimal places, shown by the bar graph. How does the rounded ratio in part (b) compare with this ratio?

  3. 99. A company that sells protective tablet cases has yearly fixed costs of $600,000. It costs the company $45 to produce each case. Each case will sell for $65. The company’s costs and revenue are modeled by the following functions, where x represents the number of tablet cases produced and sold:

    C (x)=600,000+45xThis function models thecompanys costs.R(x)=65x.This function models thecompanys revenue.

    Find and interpret (RC)(20,000), (RC)(30,000), and (RC)(40,000).

  4. 100. A department store has two locations in a city. From 2016 through 2020, the profits for each of the store’s two branches are modeled by the functions f(x)=0.44x+13.62 and g(x)=0.51x+11.14. In each model, x represents the number of years after 2016, and f and g represent the profit, in millions of dollars.

    1. What is the slope of f? Describe what this means.

    2. What is the slope of g? Describe what this means.

    3. Find f+g. What is the slope of this function? What does this mean?

  5. 101. The regular price of a laptop is x dollars. Let f(x)=x400 and g(x)=0.75x.

    1. Describe what the functions f and g model in terms of the price of the laptop.

    2. Find (fg)(x) and describe what this models in terms of the price of the laptop.

    3. Repeat part (b) for (gf)(x).

    4. Which composite function models the greater discount on the laptop, fg or gf? Explain.

  6. 102. The regular price of a pair of jeans is x dollars. Let f(x)=x5 and g(x)=0.6x.

    1. Describe what functions f and g model in terms of the price of the jeans.

    2. Find (fg)(x) and describe what this models in terms of the price of the jeans.

    3. Repeat part (b) for (gf)(x).

    4. Which composite function models the greater discount on the jeans, fg or gf? Explain.

Explaining the Concepts

  1. 103. If a function is defined by an equation, explain how to find its domain.

  2. 104. If equations for f and g are given, explain how to find fg.

  3. 105. If equations for two functions are given, explain how to obtain the quotient function and its domain.

  4. 106. Describe a procedure for finding (fg)(x). What is the name of this function?

  5. 107. Describe the values of x that must be excluded from the domain of (fg)(x).

Technology Exercises

  1. 108. Graph y1=x22x, y2=x, and y3=y1÷y2 in the same [10, 10, 1] by [10, 10, 1] viewing rectangle. Then use the TRACE feature to trace along y3. What happens at x=0? Explain why this occurs.

  2. 109. Graph y1=2x, y2=x, and y3=2y2 in the same [4, 4, 1] by [0, 2, 1] viewing rectangle. If y1 represents f and y2 represents g, use the graph of y3 to find the domain of fg. Then verify your observation algebraically.

Critical Thinking Exercises

Make Sense? In Exercises 110113, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 110. I used a function to model data from 1990 through 2020. The independent variable in my model represented the number of years after 1990, so the function’s domain was {x|x=0, 1, 2, 3,, 30}.

  2. 111. I have two functions. Function f models total world population x years after 2000 and function g models population of the world’s more-developed regions x years after 2000. I can use fg to determine the population of the world’s less-developed regions for the years in both function’s domains.

  3. 112. I must have made a mistake in finding the composite functions fg and gf, because I notice that fg is not the same function as gf.

  4. 113. This diagram illustrates that f(g(x))=x2+4.

    An illustration depicts a function f of x, x squared yields the first output, x squared, with a first input, x and function g of x, x + 4 results the second output x squared + 4 with a double input x squared.

In Exercises 114117, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 114. If f(x)=x24 and g(x)=x24, then (fg)(x)=x2 and (fg)(5)=25.

  2. 115. There can never be two functions f and g, where fg, for which (fg)(x)=(gf)(x).

  3. 116. If f(7)=5 and g(4)=7, then (fg)(4)=35.

  4. 117. If f(x)=x and g(x)=2x1, then (fg)(5)=g(2).

  5. 118. Prove that if f and g are even functions, then fg is also an even function.

  6. 119. Define two functions f and g so that fg=gf.

Retaining the Concepts

  1. 120. Solve and check: x15x+32=1x4.

    (Section P.7, Example 2)

  2. 121. In July 2020, the toll for the Golden Gate Bridge was $8.40 for drivers making one-time payments for each crossing. Drivers who pay a one-time fee of $20.00 to purchase a FasTrak toll tag pay $7.70 for each crossing. How many times must a driver cross the Golden Gate Bridge for the cost of these two options to be the same? Round to the nearest whole number. Find the total cost of each option for the rounded number of crossings.

    (Section P.8, Example 3)

  3. 122. Solve for y: Ax+By=Cy+D.

    (Section P.7, Example 5)

Preview Exercises

Exercises 123125 will help you prepare for the material covered in the next section.

  1. 123. Consider the function defined by

    {(2, 4), (1, 1), (1, 1), (2, 4)}.

    Reverse the components of each ordered pair and write the resulting relation. Is this relation a function?

  2. 124. Solve for y:x=5y+4.

  3. 125. Solve for y:x=y21, y0.

Section 1.8: Inverse Functions

Section 1.8 Inverse Functions

Learning Objectives

What You’ll Learn

  1. 1 Verify inverse functions.

  2. 2 Find the inverse of a function.

  3. 3 Use the horizontal line test to determine if a function has an inverse function.

  4. 4 Use the graph of a one-to-one function to graph its inverse function.

  5. 5 Find the inverse of a function and graph both functions on the same axes.

Based on Shakespeare’s Romeo and Juliet, the film West Side Story swept the 1961 Academy Awards with ten Oscars. The top four movies to win the most Oscars are shown in Table 1.6.

Table 1.6 Films Winning the Most Oscars

Movie Year Number of Academy Awards
Ben-Hur 1960 11
Titanic 1998 11
The Lord of the Rings: The Return of the King 2003 11
West Side Story 1961 10

Source: Russell Ash, The Top 10 of Everything, 2011

We can use the information in Table 1.6 to define a function. Let the domain of the function be the set of four movies shown in the table. Let the range be the number of Academy Awards for each of the respective films. The function can be written as follows:

f: {(Ben-Hur, 11), (Titanic, 11), (The Lord of the Rings, 11), (West Side Story, 10)}.

Now let’s “undo” f by interchanging the first and second components in each of the ordered pairs. Switching the inputs and outputs of f, we obtain the following relation:

The image shows the switching of the inputs and outputs of f.
1.8-300 Full Alternative Text

Can you see that this relation is not a function? Three of its ordered pairs have the same first component and different second components. This violates the definition of a function.

If a function f is a set of ordered pairs, (x, y) then the changes produced by f can be “undone” by reversing the components of all the ordered pairs. The resulting relation, (y, x), may or may not be a function. In this section, we will develop these ideas by studying functions whose compositions have a special “undoing” relationship.

Inverse Functions

Here are two functions that describe situations related to the price of a computer, x:

f(x)=x300g(x)=x+300.

Function f subtracts $300 from the computer’s price and function g adds $300 to the computer’s price. Let’s see what f(g(x)) does. Put g(x) into f:

The image shows the process to solve the function f of x = x minus 300.

Using f(x)=x300 and g(x)=x+300, we see that f(g(x))=x. By putting g(x) into f and finding f(g(x)), the computer’s price, x, went through two changes: the first, an increase; the second, a decrease:

x+300300.

The final price of the computer, x, is identical to its starting price, x.

In general, if the changes made to x by a function g are undone by the changes made by a function f, then

f(g(x))=x.

Assume, also, that this “undoing” takes place in the other direction:

g(f(x))=x.

Under these conditions, we say that each function is the inverse function of the other. The fact that g is the inverse of f is expressed by renaming g as f1, read “f-inverse.” For example, the inverse functions

f(x)=x300g(x)=x+300

are usually named as follows:

f(x)=x300f1(x)=x+300.

We can use partial tables of coordinates for f and f1 to gain numerical insight into the relationship between a function and its inverse function.

The image shows a table for functions f and inverse of f.

The tables illustrate that if a function f is the set of ordered pairs (x, y), then its inverse, f1, is the set of ordered pairs (y, x). Using these tables, we can see how one function’s changes to x are undone by the other function:

The inverse of f composed with f of 1300 = inverse of f left parenthesis f of 1300 right parenthesis = inverse of f of 1000 = 1300.

The final price of the computer, $1300, is identical to its starting price, $1300.

With these ideas in mind, we present the formal definition of the inverse of a function:

Definition of the Inverse of a Function

Let f and g be two functions such that

f(g(x))=xfor everyx in the domain ofg

and

g(f(x))=xfor everyx in the domain off.

The function g is the inverse of the function f and is denoted by f1 (read “f-inverse”). Thus, f(f1(x))=x and f1(f(x))=x. The domain of f is equal to the range of f1, and vice versa.

Objective 1: Verify inverse functions

  1. Objective 1 Verify inverse functions.

Watch Video

Example 1 Verifying Inverse Functions

Show that each function is the inverse of the other:

f(x)=3x+2andg(x)=x23.

Solution

To show that f and g are inverses of each other, we must show that f(g(x))=x and g(f(x))=x. We begin with f(g(x)).

The process to solve the function f of x = 3x + 2.

Next, we find g(f(x)).

The process to solve the function g of x = start fraction x minus 2 over 3 end fraction.

Because g is the inverse of f (and vice versa), we can use inverse notation and write

f(x)=3x+2andf1(x)=x23.

Notice how f1 undoes the changes produced by f: f changes x by multiplying by 3 and adding 2, and f1 undoes this by subtracting 2 and dividing by 3. This “undoing” process is illustrated in Figure 1.70.

Figure 1.70 f1 undoes the changes produced by f.

The image shows two functions that illustrate how the two functions f and g reverse the effect of each other.

Check Point 1

  • Show that each function is the inverse of the other:

    f(x)=4x7andg(x)=x+74.
Section 1.8: Inverse Functions

Section 1.8 Inverse Functions

Learning Objectives

What You’ll Learn

  1. 1 Verify inverse functions.

  2. 2 Find the inverse of a function.

  3. 3 Use the horizontal line test to determine if a function has an inverse function.

  4. 4 Use the graph of a one-to-one function to graph its inverse function.

  5. 5 Find the inverse of a function and graph both functions on the same axes.

Based on Shakespeare’s Romeo and Juliet, the film West Side Story swept the 1961 Academy Awards with ten Oscars. The top four movies to win the most Oscars are shown in Table 1.6.

Table 1.6 Films Winning the Most Oscars

Movie Year Number of Academy Awards
Ben-Hur 1960 11
Titanic 1998 11
The Lord of the Rings: The Return of the King 2003 11
West Side Story 1961 10

Source: Russell Ash, The Top 10 of Everything, 2011

We can use the information in Table 1.6 to define a function. Let the domain of the function be the set of four movies shown in the table. Let the range be the number of Academy Awards for each of the respective films. The function can be written as follows:

f: {(Ben-Hur, 11), (Titanic, 11), (The Lord of the Rings, 11), (West Side Story, 10)}.

Now let’s “undo” f by interchanging the first and second components in each of the ordered pairs. Switching the inputs and outputs of f, we obtain the following relation:

The image shows the switching of the inputs and outputs of f.
1.8-300 Full Alternative Text

Can you see that this relation is not a function? Three of its ordered pairs have the same first component and different second components. This violates the definition of a function.

If a function f is a set of ordered pairs, (x, y) then the changes produced by f can be “undone” by reversing the components of all the ordered pairs. The resulting relation, (y, x), may or may not be a function. In this section, we will develop these ideas by studying functions whose compositions have a special “undoing” relationship.

Inverse Functions

Here are two functions that describe situations related to the price of a computer, x:

f(x)=x300g(x)=x+300.

Function f subtracts $300 from the computer’s price and function g adds $300 to the computer’s price. Let’s see what f(g(x)) does. Put g(x) into f:

The image shows the process to solve the function f of x = x minus 300.

Using f(x)=x300 and g(x)=x+300, we see that f(g(x))=x. By putting g(x) into f and finding f(g(x)), the computer’s price, x, went through two changes: the first, an increase; the second, a decrease:

x+300300.

The final price of the computer, x, is identical to its starting price, x.

In general, if the changes made to x by a function g are undone by the changes made by a function f, then

f(g(x))=x.

Assume, also, that this “undoing” takes place in the other direction:

g(f(x))=x.

Under these conditions, we say that each function is the inverse function of the other. The fact that g is the inverse of f is expressed by renaming g as f1, read “f-inverse.” For example, the inverse functions

f(x)=x300g(x)=x+300

are usually named as follows:

f(x)=x300f1(x)=x+300.

We can use partial tables of coordinates for f and f1 to gain numerical insight into the relationship between a function and its inverse function.

The image shows a table for functions f and inverse of f.

The tables illustrate that if a function f is the set of ordered pairs (x, y), then its inverse, f1, is the set of ordered pairs (y, x). Using these tables, we can see how one function’s changes to x are undone by the other function:

The inverse of f composed with f of 1300 = inverse of f left parenthesis f of 1300 right parenthesis = inverse of f of 1000 = 1300.

The final price of the computer, $1300, is identical to its starting price, $1300.

With these ideas in mind, we present the formal definition of the inverse of a function:

Definition of the Inverse of a Function

Let f and g be two functions such that

f(g(x))=xfor everyx in the domain ofg

and

g(f(x))=xfor everyx in the domain off.

The function g is the inverse of the function f and is denoted by f1 (read “f-inverse”). Thus, f(f1(x))=x and f1(f(x))=x. The domain of f is equal to the range of f1, and vice versa.

Objective 2: Find the inverse of a function

Finding the Inverse of a Function

  1. Objective 2 Find the inverse of a function.

Watch Video

The definition of the inverse of a function tells us that the domain of f is equal to the range of f1, and vice versa. This means that if the function f is the set of ordered pairs (x, y), then the inverse of f is the set of ordered pairs (y, x). If a function is defined by an equation, we can obtain the equation for f1, the inverse of f, by interchanging the role of x and y in the equation for the function f.

Finding the Inverse of a Function

The equation for the inverse of a function f can be found as follows:

  1. Replace f(x) with y in the equation for f(x).

  2. Interchange x and y.

  3. Solve for y. If this equation does not define y as a function of x, the function f does not have an inverse function and this procedure ends. If this equation does define y as a function of x, the function f has an inverse function.

  4. If f has an inverse function, replace y in step 3 by f1(x). We can verify our result by showing that f(f1(x))=x and f1(f(x))=x.

The procedure for finding a function’s inverse uses a switch-and-solve strategy. Switch x and y, and then solve for y.

Example 2 Finding the Inverse of a Function

Find the inverse of f(x)=7x5.

Solution

  1. Step 1 REPLACE f(x) WITH y:

    y=7x5.
  2. Step 2 INTERCHANGE x AND y:

    x=7y5.This is the inverse function.
  3. Step 3 SOLVE FOR y:

    x+5=7yAdd 5 to both sides.x+57=y.Divide both sides by 7.
  4. Step 4 REPLACE y WITH f 1(x):

    f1(x)=x+57.The equation is writtenwith f1 on the left.

Thus, the inverse of f(x)=7x5 is f1(x)=x+57.

The inverse function, f1, undoes the changes produced by f.f changes x by multiplying by 7 and subtracting 5. f1 undoes this by adding 5 and dividing by 7.

Check Point 2

  • Find the inverse of f(x)=2x+7.

Example 3 Finding the Inverse of a Function

Find the inverse of f(x)=x3+1.

Solution

  1. Step 1 REPLACE f(x) WITH y: y=x3+1.

  2. Step 2 INTERCHANGE x AND y: x=y3+1.

  3. Step 3 SOLVE FOR y:

    The image shows the process to find the inverse of a function.
  4. Step 4 REPLACE y WITH f 1(x): f1(x)=x13.

Thus, the inverse of f(x)=x3+1 is f1(x)=x13.

Check Point 3

  • Find the inverse of f(x)=4x31.

Example 4 Finding the Inverse of a Function

Find the inverse of f(x)=x+2x3,x3.

Solution

  1. Step 1 REPLACE f(x) with y:

    y=x+2x3.
  2. Step 2 INTERCHANGE x AND y

    x=y+2y3.
  3. Step 3 SOLVE FOR y:

    The image shows a mathematical expression to find the value of y.

  4. Step 4 REPLACE y WITH f1(x):

    f1(x)=3x+2x1.

Thus, the inverse of f(x)=x+2x3 is f1(x)=3x+2x1.

Check Point 4

  • Find the inverse of f(x)=x+1x5,x5.

Objective 2: Find the inverse of a function

Finding the Inverse of a Function

  1. Objective 2 Find the inverse of a function.

Watch Video

The definition of the inverse of a function tells us that the domain of f is equal to the range of f1, and vice versa. This means that if the function f is the set of ordered pairs (x, y), then the inverse of f is the set of ordered pairs (y, x). If a function is defined by an equation, we can obtain the equation for f1, the inverse of f, by interchanging the role of x and y in the equation for the function f.

Finding the Inverse of a Function

The equation for the inverse of a function f can be found as follows:

  1. Replace f(x) with y in the equation for f(x).

  2. Interchange x and y.

  3. Solve for y. If this equation does not define y as a function of x, the function f does not have an inverse function and this procedure ends. If this equation does define y as a function of x, the function f has an inverse function.

  4. If f has an inverse function, replace y in step 3 by f1(x). We can verify our result by showing that f(f1(x))=x and f1(f(x))=x.

The procedure for finding a function’s inverse uses a switch-and-solve strategy. Switch x and y, and then solve for y.

Example 2 Finding the Inverse of a Function

Find the inverse of f(x)=7x5.

Solution

  1. Step 1 REPLACE f(x) WITH y:

    y=7x5.
  2. Step 2 INTERCHANGE x AND y:

    x=7y5.This is the inverse function.
  3. Step 3 SOLVE FOR y:

    x+5=7yAdd 5 to both sides.x+57=y.Divide both sides by 7.
  4. Step 4 REPLACE y WITH f 1(x):

    f1(x)=x+57.The equation is writtenwith f1 on the left.

Thus, the inverse of f(x)=7x5 is f1(x)=x+57.

The inverse function, f1, undoes the changes produced by f.f changes x by multiplying by 7 and subtracting 5. f1 undoes this by adding 5 and dividing by 7.

Check Point 2

  • Find the inverse of f(x)=2x+7.

Example 3 Finding the Inverse of a Function

Find the inverse of f(x)=x3+1.

Solution

  1. Step 1 REPLACE f(x) WITH y: y=x3+1.

  2. Step 2 INTERCHANGE x AND y: x=y3+1.

  3. Step 3 SOLVE FOR y:

    The image shows the process to find the inverse of a function.
  4. Step 4 REPLACE y WITH f 1(x): f1(x)=x13.

Thus, the inverse of f(x)=x3+1 is f1(x)=x13.

Check Point 3

  • Find the inverse of f(x)=4x31.

Example 4 Finding the Inverse of a Function

Find the inverse of f(x)=x+2x3,x3.

Solution

  1. Step 1 REPLACE f(x) with y:

    y=x+2x3.
  2. Step 2 INTERCHANGE x AND y

    x=y+2y3.
  3. Step 3 SOLVE FOR y:

    The image shows a mathematical expression to find the value of y.

  4. Step 4 REPLACE y WITH f1(x):

    f1(x)=3x+2x1.

Thus, the inverse of f(x)=x+2x3 is f1(x)=3x+2x1.

Check Point 4

  • Find the inverse of f(x)=x+1x5,x5.

Objective 2: Find the inverse of a function

Finding the Inverse of a Function

  1. Objective 2 Find the inverse of a function.

Watch Video

The definition of the inverse of a function tells us that the domain of f is equal to the range of f1, and vice versa. This means that if the function f is the set of ordered pairs (x, y), then the inverse of f is the set of ordered pairs (y, x). If a function is defined by an equation, we can obtain the equation for f1, the inverse of f, by interchanging the role of x and y in the equation for the function f.

Finding the Inverse of a Function

The equation for the inverse of a function f can be found as follows:

  1. Replace f(x) with y in the equation for f(x).

  2. Interchange x and y.

  3. Solve for y. If this equation does not define y as a function of x, the function f does not have an inverse function and this procedure ends. If this equation does define y as a function of x, the function f has an inverse function.

  4. If f has an inverse function, replace y in step 3 by f1(x). We can verify our result by showing that f(f1(x))=x and f1(f(x))=x.

The procedure for finding a function’s inverse uses a switch-and-solve strategy. Switch x and y, and then solve for y.

Example 2 Finding the Inverse of a Function

Find the inverse of f(x)=7x5.

Solution

  1. Step 1 REPLACE f(x) WITH y:

    y=7x5.
  2. Step 2 INTERCHANGE x AND y:

    x=7y5.This is the inverse function.
  3. Step 3 SOLVE FOR y:

    x+5=7yAdd 5 to both sides.x+57=y.Divide both sides by 7.
  4. Step 4 REPLACE y WITH f 1(x):

    f1(x)=x+57.The equation is writtenwith f1 on the left.

Thus, the inverse of f(x)=7x5 is f1(x)=x+57.

The inverse function, f1, undoes the changes produced by f.f changes x by multiplying by 7 and subtracting 5. f1 undoes this by adding 5 and dividing by 7.

Check Point 2

  • Find the inverse of f(x)=2x+7.

Example 3 Finding the Inverse of a Function

Find the inverse of f(x)=x3+1.

Solution

  1. Step 1 REPLACE f(x) WITH y: y=x3+1.

  2. Step 2 INTERCHANGE x AND y: x=y3+1.

  3. Step 3 SOLVE FOR y:

    The image shows the process to find the inverse of a function.
  4. Step 4 REPLACE y WITH f 1(x): f1(x)=x13.

Thus, the inverse of f(x)=x3+1 is f1(x)=x13.

Check Point 3

  • Find the inverse of f(x)=4x31.

Example 4 Finding the Inverse of a Function

Find the inverse of f(x)=x+2x3,x3.

Solution

  1. Step 1 REPLACE f(x) with y:

    y=x+2x3.
  2. Step 2 INTERCHANGE x AND y

    x=y+2y3.
  3. Step 3 SOLVE FOR y:

    The image shows a mathematical expression to find the value of y.

  4. Step 4 REPLACE y WITH f1(x):

    f1(x)=3x+2x1.

Thus, the inverse of f(x)=x+2x3 is f1(x)=3x+2x1.

Check Point 4

  • Find the inverse of f(x)=x+1x5,x5.

Objective 3: Use the horizontal line test to determine if a function has an inverse function

The Horizontal Line Test and One-to-One Functions

  1. Objective 3 Use the horizontal line test to determine if a function has an inverse function.

Watch Video

Let’s see what happens if we try to find the inverse of the standard quadratic function, f(x)=x2.

  1. Step 1 Replace f(x) with y: y=x2.

  2. Step 2 Interchange x and y: x=y2.

  3. Step 3 Solve for y: We apply the square root property to solve y2=x for y. We obtain

    y=±x.

The ± in y=±x shows that for certain values of x (all positive real numbers), there are two values of y. Because this equation does not represent y as a function of x, the standard quadratic function f(x)=x2 does not have an inverse function.

We can use a few of the solutions of y=x2 to illustrate numerically that this function does not have an inverse:

Four solutions of y = x squared.
1.8-309 Full Alternative Text

A function provides exactly one output for each input. Thus, the ordered pairs in the bottom row do not define a function.

Can we look at the graph of a function and tell if it represents a function with an inverse? Yes. The graph of the standard quadratic function f(x)=x2 is shown in Figure 1.71. Four units above the x-axis, a horizontal line is drawn. This line intersects the graph at two of its points, (2, 4) and (2, 4). Inverse functions have ordered pairs with the coordinates reversed. We just saw what happened when we interchanged x and y. We obtained (4, 2) and (4, 2), and these ordered pairs do not define a function.

Figure 1.71 The horizontal line intersects the graph twice.

The graph for f of x = x squared is an upward opening parabola that falls through (negative 2, 4) with its vertex at the (0, 0) and again rises through (2, 4). A horizontal line intersects the parabola at (negative 2, 4) and (2, 4).

If any horizontal line, such as the one in Figure 1.71, intersects a graph at two or more points, the set of these points will not define a function when their coordinates are reversed. This suggests the horizontal line test for inverse functions.

The Horizontal Line Test for Inverse Functions

A function f has an inverse that is a function, f1, if there is no horizontal line that intersects the graph of the function f at more than one point.

Example 5 Applying the Horizontal Line Test

Which of the following graphs represent functions that have inverse functions?

The image shows four graphs.

Solution

Notice that horizontal lines can be drawn in graphs (b) and (c) that intersect the graphs more than once. These graphs do not pass the horizontal line test. These are not the graphs of functions with inverse functions. By contrast, no horizontal line can be drawn in graphs (a) and (d) that intersects the graphs more than once. These graphs pass the horizontal line test. Thus, the graphs in parts (a) and (d) represent functions that have inverse functions.

The image shows four graphs.

Check Point 5

A function passes the horizontal line test when no two different ordered pairs have the same second component. This means that if x1x2, then f(x1)f(x2). Such a function is called a one-to-one function. Thus, a one-to-one function is a function in which no two different ordered pairs have the same second component. Only one-to-one functions have inverse functions. Any function that passes the horizontal line test is a one-to-one function. Any one-to-one function has a graph that passes the horizontal line test.

Objective 4: Use the graph of a one-to-one function to graph its inverse function

Graphs of f and f1

  1. Objective 4 Use the graph of a one-to-one function to graph its inverse function.

Watch Video

There is a relationship between the graph of a one-to-one function, f, and its inverse, f1. Because inverse functions have ordered pairs with the coordinates interchanged, if the point (a, b) is on the graph of f, then the point (b, a) is on the graph of f1. The points (a, b) and (b, a) are symmetric with respect to the line y=x. Thus, the graph of f 1 is a reflection of the graph of f about the line y=x. This is illustrated in Figure 1.72.

Figure 1.72 The graph of f1 is a reflection of the graph of f about y=x.

The image shows a graph of inverse function displays two curves and a line.
Figure 1.72 Full Alternative Text

Example 6 Graphing the Inverse Function

Use the graph of f in Figure 1.73 to draw the graph of its inverse function.

Figure 1.73

The graph for y = f of x is a line that starts from (negative 3, negative 2) passes through (negative 1, 0), and ends at (4, 2).

Solution

We begin by noting that no horizontal line intersects the graph of f at more than one point, so f does have an inverse function. Because the points (3, 2), (1, 0), and (4, 2) are on the graph of f, the graph of the inverse function, f1, has points with these ordered pairs reversed. Thus, (2, 3), (0, 1), and (2, 4) are on the graph of f1. We can use these points to graph f1. The graph of f1 is shown in green in Figure 1.74. Note that the green graph of f1 is the reflection of the blue graph of f about the line y=x.

Figure 1.74 The graphs of f and f1

The image shows a graph of inverse function displays three lines.

Check Point 6

  • The graph of function f consists of two line segments, one segment from (2, 2) to (1, 0) and a second segment from (1, 0) to (1, 2). Graph f and use the graph to draw the graph of its inverse function.

Objective 5: Find the inverse of a function and graph both functions on the same axes

  1. Objective 5 Find the inverse of a function and graph both functions on the same axes.

Watch Video

In our final example, we will first find f1. Then we will graph f and f1 in the same rectangular coordinate system.

Example 7 Finding the Inverse of a Domain-Restricted Function

Find the inverse of f(x)=x21 if x0. Graph f and f1 in the same rectangular coordinate system.

Solution

The graph of f(x)=x21 is the graph of the standard quadratic function shifted vertically down 1 unit. Figure 1.75 shows the function’s graph. This graph fails the horizontal line test, so the function f(x)=x21 does not have an inverse function. By restricting the domain to x0, as given, we obtain a new function whose graph is shown in red in Figure 1.75. This red portion of the graph is increasing on the interval (0, ) and passes the horizontal line test. This tells us that f(x)=x21 has an inverse function if we restrict its domain to x0. We use our four-step procedure to find this inverse function. Begin with f(x)=x21, x0.

Figure 1.75

The graph for f of x = x squared minus 1 is an upward opening parabola with indefinite ends that falls through (negative 2, 3) with its vertex at (0, negative 1) and again rises through (2, 3).
  1. Step 1 REPLACE f(x) WITH y: y=x21, x0.

  2. Step 2 INTERCHANGE x AND y: x=y21, y0.

  3. Step 3 SOLVE FOR y:

    The image shows the process to find the inverse of a function.
  4. Step 4 REPLACE y WITH f 1(x): f1(x)=x+1.

    Thus, the inverse of f(x)=x21, x0, is f1(x)=x+1. The graphs of f and f1 are shown in Figure 1.76. We obtained the graph of f1(x)=x+1 by shifting the graph of the square root function, y=x, horizontally to the left 1 unit. Note that the green graph of f1 is the reflection of the red graph of f about the line y=x.

Figure 1.76

The image shows a graph of inverse function displays two curves and a line.

Check Point 7

  • Find the inverse of f(x)=x2+1 if x0. Graph f and f1 in the same rectangular coordinate system.

1.8: Exercise Set

1.8 Exercise Set

Practice Exercises

In Exercises 110, find f(g(x)) and g(f(x)) and determine whether each pair of functions f and g are inverses of each other.

  1. 1. f(x)=4x and g(x)=x4

  2. 2. f(x)=6x and g(x)=x6

  3. 3. f(x)=3x+8 and g(x)=x83

  4. 4. f(x)=4x+9 and g(x)=x94

  5. 5. f(x)=5x9 and g(x)=x+59

  6. 6. f(x)=3x7 and g(x)=x+37

  7. 7. f(x)=3x4 and g(x)=3x+4

  8. 8. f(x)=2x5 and g(x)=2x+5

  9. 9. f(x)=x and g(x)=x

  10. 10. f(x)=x43 and g(x)=x3+4

The functions in Exercises 1128 are all one-to-one. For each function,

  1. Find an equation for f1(x), the inverse function.

  2. Verify that your equation is correct by showing that f(f1(x))=x and f1(f(x))=x.

  1. 11. f(x)=x+3

  2. 12. f(x)=x+5

  3. 13. f(x)=2x

  4. 14. f(x)=4x

  5. 15. f(x)=2x+3

  6. 16. f(x)=3x1

  7. 17. f(x)=x3+2

  8. 18. f(x)=x31

  9. 19. f(x)=(x+2)3

  10. 20. f(x)=(x1)3

  11. 21. f(x)=1x

  12. 22. f(x)=2x

  13. 23. f(x)=x

  14. 24. f(x)=x3

  15. 25. f(x)=x+4x2

  16. 26. f(x)=x+5x6

  17. 27. f(x)=2x+1x3

  18. 28. f(x)=2x3x+1

Which graphs in Exercises 2934 represent functions that have inverse functions?

  1. 29.

    A graph plots a semicircle that starts from the second quadrant and ends at the first quadrant.
  2. 30.

    A graph plots a curve that concaves down increases in the third quadrant passes through the origin and concave up increases in the first quadrant.
  3. 31.

    A graph plots a curve that falls through the negative x axis to the local minimum in the third quadrant. The curve then rises through the origin to the local maximum in the first quadrant and falls to the fourth quadrant through the positive x axis.
  4. 32.

    A graph is a line that rises from a point in the negative x axis and passes through the second quadrant and the first quadrant. After a certain point in the first quadrant, it runs parallel to the x axis.

  5. 33.

    A graph plots a concave up an increasing curve that starts from the second quadrant, passes through positive y intercept, and extends in the first quadrant.
  6. 34.

    A graph plots a concave down increasing curve starts that from the fourth quadrant, passes through positive x intercept, and extends in the first quadrant.

In Exercises 3538, use the graph of f to draw the graph of its inverse function.

  1. 35.

    A graph plots a line that rises through (0, negative 4), (2, 0), (3, 2), and (4, 4).
  2. 36.

    A graph plots a line that starts from (negative 3, 0), passes through (negative 1, 2), and (3, 3).
  3. 37.

    A graph plots a concave up an increasing curve that passes through (negative 4, 0), (0, 1), (1, 2), and (2, 4).
  4. 38.

    A graph plots a concave up decreasing curve that passes through (negative 2, 4), (negative 1, 2), (0, 1), and (4, 0).

In Exercises 3952,

  1. Find an equation for f1(x).

  2. Graph f and f1 in the same rectangular coordinate system.

  3. Use interval notation to give the domain and the range of f and f1.

  1. 39. f(x)=2x1

  2. 40. f(x)=2x3

  3. 41. f(x)=x24, x0

  4. 42. f(x)=x21, x0

  5. 43. f(x)=(x1)2, x1

  6. 44. f(x)=(x1)2, x1

  7. 45. f(x)=x31

  8. 46. f(x)=x3+1

  9. 47. f(x)=(x+2)3

  10. 48. f(x)=(x2)3

(Hint for Exercises 4952: To solve for a variable involving an nth root, raise both sides of the equation to the nth power: (yn)n=y.)

  1. 49. f(x)=x1

  2. 50. f(x)=x+2

  3. 51. f(x)=x3+1

  4. 52. f(x)=x13

Practice PLUS

In Exercises 5358, f and g are defined by the following tables. Use the tables to evaluate each composite function.

x f(x)
1 1
0 4
1 5
2 1
x g(x)
1 0
1 1
4 2
10 1
  1. 53. f(g(1))

  2. 54. f(g(4))

  3. 55. (gf)(1)

  4. 56. (gf)(0)

  5. 57. f1(g(10))

  6. 58. f1(g(1))

In Exercises 5964, let

f(x)=2x5g(x)=4x1h(x)=x2+x+2.

Evaluate the indicated function without finding an equation for the function.

  1. 59. (fg)(0)

  2. 60. (gf)(0)

  3. 61. f1(1)

  4. 62. g1(7)

  5. 63. g(f[h(1)])

  6. 64. f(g[h(1)])

Application Exercises

Way to Go Holland was the first country to establish an official bicycle policy. It currently has over 12,000 miles of paths and lanes exclusively for bicycles. The graph shows the percentage of travel by bike and by car in Holland, as well as in four other selected countries. Use the information in the graph to solve Exercises 6566.

The image shows a bar graph titled, modes of travel in selected countries by bike and by car.

Source: EUROSTAT

  1. 65.

    1. Consider a function, f, whose domain is the set of the five countries shown in the graph. Let the range be the percentage of travel by bike in each of the respective countries. Write the function f as a set of ordered pairs.

    2. Write the relation that is the inverse of f as a set of ordered pairs. Is this relation a function? Explain your answer.

  2. 66.

    1. Consider a function, f, whose domain is the set of the five countries shown in the graph. Let the range be the percentage of travel by car in each of the respective countries. Write the function f as a set of ordered pairs.

    2. Write the relation that is the inverse of f as a set of ordered pairs. Is this relation a function? Explain your answer.

  3. 67. The graph represents the probability of two people in the same room sharing a birthday as a function of the number of people in the room. Call the function f.

    A graph plots probability that two people share the same birthday versus the number of persons. The horizontal axis ranges from 0 to 60 in increments of 10, and the vertical axis ranges from 0.1 to 1.0 in increments of 0.1.
    1. Explain why f has an inverse that is a function.

    2. Describe in practical terms the meaning of f1(0.25),f1(0.5), and f1(0.7).

  4. 68. A study of 900 working women in Texas showed that their feelings changed throughout the day. As the graph indicates, the women felt better as time passed, except for a blip (that’s slang for relative maximum) at lunchtime.

    The image shows a line graph titled, average level of happiness at different times a day plots the average level of happiness versus time of day.

    Source: D. Kahneman et al., “A Survey Method for Characterizing Daily Life Experience,” Science

    1. Does the graph have an inverse that is a function? Explain your answer.

    2. Identify two or more times of day when the average happiness level is 3. Express your answers as ordered pairs.

    3. Do the ordered pairs in part (b) indicate that the graph represents a one-to-one function? Explain your answer.

  5. 69. The formula

    y=f(x)=95 x+32

    is used to convert from x degrees Celsius to y degrees Fahrenheit. The formula

    y=g(x)=59 (x32)

    is used to convert from x degrees Fahrenheit to y degrees Celsius. Show that f and g are inverse functions.

Explaining the Concepts

  1. 70. Explain how to determine if two functions are inverses of each other.

  2. 71. Describe how to find the inverse of a one-to-one function.

  3. 72. What is the horizontal line test and what does it indicate?

  4. 73. Describe how to use the graph of a one-to-one function to draw the graph of its inverse function.

  5. 74. How can a graphing utility be used to visually determine if two functions are inverses of each other?

  6. 75. What explanations can you offer for the trends shown by the graph in Exercise 68?

Technology Exercises

In Exercises 7683, use a graphing utility to graph the function. Use the graph to determine whether the function has an inverse that is a function (that is, whether the function is one-to-one).

  1. 76. f(x)=x21

  2. 77. f(x)=2x3

  3. 78. f(x)=x32

  4. 79. f(x)=x44

  5. 80. f(x)=int(x2)

  6. 81. f(x)=|x2|

  7. 82. f(x)=(x1)3

  8. 83. f(x)=16x2

In Exercises 8486, use a graphing utility to graph f and g in the same [8, 8, 1] by [5, 5, 1] viewing rectangle. In addition, graph the line y=x and visually determine if f and g are inverses.

  1. 84. f(x)=4x+4, g(x)=0.25x1

  2. 85. f(x)=1x+2, g(x)=1x2

  3. 86. f(x)=x32, g(x)=(x+2)3

Critical Thinking Exercises

Make Sense? In Exercises 8790, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 87. I found the inverse of f(x)=5x4 in my head: The reverse of multiplying by 5 and subtracting 4 is adding 4 and dividing by 5, so f1(x)=x+45.

  2. 88. I’m working with the linear function f(x)=3x+5 and I do not need to find f1 in order to determine the value of (ff1)(17).

  3. 89. When finding the inverse of a function, I interchange x and y, which reverses the domain and range between the function and its inverse.

  4. 90. I used vertical lines to determine if my graph represents a one-to-one function.

In Exercises 9194, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 91. The inverse of {(1, 4), (2, 7)} is {(2, 7), (1, 4)}.

  2. 92. The function f(x)=5 is one-to-one.

  3. 93. If f(x)=3x, then f1(x)=13x.

  4. 94. The domain of f is the same as the range of f1.

  5. 95. If f(x)=3x and g(x)=x+5, find (fg)1(x) and (g1f1)(x).

  6. 96. Show that

    f(x)=3x25x3

    is its own inverse.

  7. 97. Freedom 7 was the spacecraft that carried the first American into space in 1961. Total flight time was 15 minutes and the spacecraft reached a maximum height of 116 miles. Consider a function, s, that expresses Freedom 7’s height, s(t), in miles, after t minutes. Is s a one-to-one function? Explain your answer.

  8. 98. If f(2)=6, and f is one-to-one, find x satisfying 8+f1(x1)=10.

Group Exercise

  1. 99. In Tom Stoppard’s play Arcadia, the characters dream and talk about mathematics, including ideas involving graphing, composite functions, symmetry, and lack of symmetry in things that are tangled, mysterious, and unpredictable. Group members should read the play. Present a report on the ideas discussed by the characters that are related to concepts that we studied in this chapter. Bring in a copy of the play and read appropriate excerpts.

Retaining the Concepts

  1. 100. Solve by completing the square:

    2x25x+1=0.

    (Section P.7, Example 9)

  2. 101. The size of a television screen refers to the length of its diagonal. If the length of an HDTV screen is 28 inches and its width is 15.7 inches, what is the size of the screen to the nearest inch? (Section P.8, Example 8)

  3. 102. Solve and graph the solution set on a number line:

    3|2x1|21.

    (Section P.9, Example 9)

Preview Exercises

Exercises 103105 will help you prepare for the material covered in the next section.

  1. 103. Let (x1, y1)=(7, 2) and (x2, y2)=(1, 1). Find (x2x1)2+(y2y1)2. Express the answer in simplified radical form.

  2. 104. Use a rectangular coordinate system to graph the circle with center (1, 1) and radius 1.

  3. 105. Solve by completing the square: y26y4=0.

1.8: Exercise Set

1.8 Exercise Set

Practice Exercises

In Exercises 110, find f(g(x)) and g(f(x)) and determine whether each pair of functions f and g are inverses of each other.

  1. 1. f(x)=4x and g(x)=x4

  2. 2. f(x)=6x and g(x)=x6

  3. 3. f(x)=3x+8 and g(x)=x83

  4. 4. f(x)=4x+9 and g(x)=x94

  5. 5. f(x)=5x9 and g(x)=x+59

  6. 6. f(x)=3x7 and g(x)=x+37

  7. 7. f(x)=3x4 and g(x)=3x+4

  8. 8. f(x)=2x5 and g(x)=2x+5

  9. 9. f(x)=x and g(x)=x

  10. 10. f(x)=x43 and g(x)=x3+4

The functions in Exercises 1128 are all one-to-one. For each function,

  1. Find an equation for f1(x), the inverse function.

  2. Verify that your equation is correct by showing that f(f1(x))=x and f1(f(x))=x.

  1. 11. f(x)=x+3

  2. 12. f(x)=x+5

  3. 13. f(x)=2x

  4. 14. f(x)=4x

  5. 15. f(x)=2x+3

  6. 16. f(x)=3x1

  7. 17. f(x)=x3+2

  8. 18. f(x)=x31

  9. 19. f(x)=(x+2)3

  10. 20. f(x)=(x1)3

  11. 21. f(x)=1x

  12. 22. f(x)=2x

  13. 23. f(x)=x

  14. 24. f(x)=x3

  15. 25. f(x)=x+4x2

  16. 26. f(x)=x+5x6

  17. 27. f(x)=2x+1x3

  18. 28. f(x)=2x3x+1

Which graphs in Exercises 2934 represent functions that have inverse functions?

  1. 29.

    A graph plots a semicircle that starts from the second quadrant and ends at the first quadrant.
  2. 30.

    A graph plots a curve that concaves down increases in the third quadrant passes through the origin and concave up increases in the first quadrant.
  3. 31.

    A graph plots a curve that falls through the negative x axis to the local minimum in the third quadrant. The curve then rises through the origin to the local maximum in the first quadrant and falls to the fourth quadrant through the positive x axis.
  4. 32.

    A graph is a line that rises from a point in the negative x axis and passes through the second quadrant and the first quadrant. After a certain point in the first quadrant, it runs parallel to the x axis.

  5. 33.

    A graph plots a concave up an increasing curve that starts from the second quadrant, passes through positive y intercept, and extends in the first quadrant.
  6. 34.

    A graph plots a concave down increasing curve starts that from the fourth quadrant, passes through positive x intercept, and extends in the first quadrant.

In Exercises 3538, use the graph of f to draw the graph of its inverse function.

  1. 35.

    A graph plots a line that rises through (0, negative 4), (2, 0), (3, 2), and (4, 4).
  2. 36.

    A graph plots a line that starts from (negative 3, 0), passes through (negative 1, 2), and (3, 3).
  3. 37.

    A graph plots a concave up an increasing curve that passes through (negative 4, 0), (0, 1), (1, 2), and (2, 4).
  4. 38.

    A graph plots a concave up decreasing curve that passes through (negative 2, 4), (negative 1, 2), (0, 1), and (4, 0).

In Exercises 3952,

  1. Find an equation for f1(x).

  2. Graph f and f1 in the same rectangular coordinate system.

  3. Use interval notation to give the domain and the range of f and f1.

  1. 39. f(x)=2x1

  2. 40. f(x)=2x3

  3. 41. f(x)=x24, x0

  4. 42. f(x)=x21, x0

  5. 43. f(x)=(x1)2, x1

  6. 44. f(x)=(x1)2, x1

  7. 45. f(x)=x31

  8. 46. f(x)=x3+1

  9. 47. f(x)=(x+2)3

  10. 48. f(x)=(x2)3

(Hint for Exercises 4952: To solve for a variable involving an nth root, raise both sides of the equation to the nth power: (yn)n=y.)

  1. 49. f(x)=x1

  2. 50. f(x)=x+2

  3. 51. f(x)=x3+1

  4. 52. f(x)=x13

Practice PLUS

In Exercises 5358, f and g are defined by the following tables. Use the tables to evaluate each composite function.

x f(x)
1 1
0 4
1 5
2 1
x g(x)
1 0
1 1
4 2
10 1
  1. 53. f(g(1))

  2. 54. f(g(4))

  3. 55. (gf)(1)

  4. 56. (gf)(0)

  5. 57. f1(g(10))

  6. 58. f1(g(1))

In Exercises 5964, let

f(x)=2x5g(x)=4x1h(x)=x2+x+2.

Evaluate the indicated function without finding an equation for the function.

  1. 59. (fg)(0)

  2. 60. (gf)(0)

  3. 61. f1(1)

  4. 62. g1(7)

  5. 63. g(f[h(1)])

  6. 64. f(g[h(1)])

Application Exercises

Way to Go Holland was the first country to establish an official bicycle policy. It currently has over 12,000 miles of paths and lanes exclusively for bicycles. The graph shows the percentage of travel by bike and by car in Holland, as well as in four other selected countries. Use the information in the graph to solve Exercises 6566.

The image shows a bar graph titled, modes of travel in selected countries by bike and by car.

Source: EUROSTAT

  1. 65.

    1. Consider a function, f, whose domain is the set of the five countries shown in the graph. Let the range be the percentage of travel by bike in each of the respective countries. Write the function f as a set of ordered pairs.

    2. Write the relation that is the inverse of f as a set of ordered pairs. Is this relation a function? Explain your answer.

  2. 66.

    1. Consider a function, f, whose domain is the set of the five countries shown in the graph. Let the range be the percentage of travel by car in each of the respective countries. Write the function f as a set of ordered pairs.

    2. Write the relation that is the inverse of f as a set of ordered pairs. Is this relation a function? Explain your answer.

  3. 67. The graph represents the probability of two people in the same room sharing a birthday as a function of the number of people in the room. Call the function f.

    A graph plots probability that two people share the same birthday versus the number of persons. The horizontal axis ranges from 0 to 60 in increments of 10, and the vertical axis ranges from 0.1 to 1.0 in increments of 0.1.
    1. Explain why f has an inverse that is a function.

    2. Describe in practical terms the meaning of f1(0.25),f1(0.5), and f1(0.7).

  4. 68. A study of 900 working women in Texas showed that their feelings changed throughout the day. As the graph indicates, the women felt better as time passed, except for a blip (that’s slang for relative maximum) at lunchtime.

    The image shows a line graph titled, average level of happiness at different times a day plots the average level of happiness versus time of day.

    Source: D. Kahneman et al., “A Survey Method for Characterizing Daily Life Experience,” Science

    1. Does the graph have an inverse that is a function? Explain your answer.

    2. Identify two or more times of day when the average happiness level is 3. Express your answers as ordered pairs.

    3. Do the ordered pairs in part (b) indicate that the graph represents a one-to-one function? Explain your answer.

  5. 69. The formula

    y=f(x)=95 x+32

    is used to convert from x degrees Celsius to y degrees Fahrenheit. The formula

    y=g(x)=59 (x32)

    is used to convert from x degrees Fahrenheit to y degrees Celsius. Show that f and g are inverse functions.

Explaining the Concepts

  1. 70. Explain how to determine if two functions are inverses of each other.

  2. 71. Describe how to find the inverse of a one-to-one function.

  3. 72. What is the horizontal line test and what does it indicate?

  4. 73. Describe how to use the graph of a one-to-one function to draw the graph of its inverse function.

  5. 74. How can a graphing utility be used to visually determine if two functions are inverses of each other?

  6. 75. What explanations can you offer for the trends shown by the graph in Exercise 68?

Technology Exercises

In Exercises 7683, use a graphing utility to graph the function. Use the graph to determine whether the function has an inverse that is a function (that is, whether the function is one-to-one).

  1. 76. f(x)=x21

  2. 77. f(x)=2x3

  3. 78. f(x)=x32

  4. 79. f(x)=x44

  5. 80. f(x)=int(x2)

  6. 81. f(x)=|x2|

  7. 82. f(x)=(x1)3

  8. 83. f(x)=16x2

In Exercises 8486, use a graphing utility to graph f and g in the same [8, 8, 1] by [5, 5, 1] viewing rectangle. In addition, graph the line y=x and visually determine if f and g are inverses.

  1. 84. f(x)=4x+4, g(x)=0.25x1

  2. 85. f(x)=1x+2, g(x)=1x2

  3. 86. f(x)=x32, g(x)=(x+2)3

Critical Thinking Exercises

Make Sense? In Exercises 8790, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 87. I found the inverse of f(x)=5x4 in my head: The reverse of multiplying by 5 and subtracting 4 is adding 4 and dividing by 5, so f1(x)=x+45.

  2. 88. I’m working with the linear function f(x)=3x+5 and I do not need to find f1 in order to determine the value of (ff1)(17).

  3. 89. When finding the inverse of a function, I interchange x and y, which reverses the domain and range between the function and its inverse.

  4. 90. I used vertical lines to determine if my graph represents a one-to-one function.

In Exercises 9194, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 91. The inverse of {(1, 4), (2, 7)} is {(2, 7), (1, 4)}.

  2. 92. The function f(x)=5 is one-to-one.

  3. 93. If f(x)=3x, then f1(x)=13x.

  4. 94. The domain of f is the same as the range of f1.

  5. 95. If f(x)=3x and g(x)=x+5, find (fg)1(x) and (g1f1)(x).

  6. 96. Show that

    f(x)=3x25x3

    is its own inverse.

  7. 97. Freedom 7 was the spacecraft that carried the first American into space in 1961. Total flight time was 15 minutes and the spacecraft reached a maximum height of 116 miles. Consider a function, s, that expresses Freedom 7’s height, s(t), in miles, after t minutes. Is s a one-to-one function? Explain your answer.

  8. 98. If f(2)=6, and f is one-to-one, find x satisfying 8+f1(x1)=10.

Group Exercise

  1. 99. In Tom Stoppard’s play Arcadia, the characters dream and talk about mathematics, including ideas involving graphing, composite functions, symmetry, and lack of symmetry in things that are tangled, mysterious, and unpredictable. Group members should read the play. Present a report on the ideas discussed by the characters that are related to concepts that we studied in this chapter. Bring in a copy of the play and read appropriate excerpts.

Retaining the Concepts

  1. 100. Solve by completing the square:

    2x25x+1=0.

    (Section P.7, Example 9)

  2. 101. The size of a television screen refers to the length of its diagonal. If the length of an HDTV screen is 28 inches and its width is 15.7 inches, what is the size of the screen to the nearest inch? (Section P.8, Example 8)

  3. 102. Solve and graph the solution set on a number line:

    3|2x1|21.

    (Section P.9, Example 9)

Preview Exercises

Exercises 103105 will help you prepare for the material covered in the next section.

  1. 103. Let (x1, y1)=(7, 2) and (x2, y2)=(1, 1). Find (x2x1)2+(y2y1)2. Express the answer in simplified radical form.

  2. 104. Use a rectangular coordinate system to graph the circle with center (1, 1) and radius 1.

  3. 105. Solve by completing the square: y26y4=0.

1.8: Exercise Set

1.8 Exercise Set

Practice Exercises

In Exercises 110, find f(g(x)) and g(f(x)) and determine whether each pair of functions f and g are inverses of each other.

  1. 1. f(x)=4x and g(x)=x4

  2. 2. f(x)=6x and g(x)=x6

  3. 3. f(x)=3x+8 and g(x)=x83

  4. 4. f(x)=4x+9 and g(x)=x94

  5. 5. f(x)=5x9 and g(x)=x+59

  6. 6. f(x)=3x7 and g(x)=x+37

  7. 7. f(x)=3x4 and g(x)=3x+4

  8. 8. f(x)=2x5 and g(x)=2x+5

  9. 9. f(x)=x and g(x)=x

  10. 10. f(x)=x43 and g(x)=x3+4

The functions in Exercises 1128 are all one-to-one. For each function,

  1. Find an equation for f1(x), the inverse function.

  2. Verify that your equation is correct by showing that f(f1(x))=x and f1(f(x))=x.

  1. 11. f(x)=x+3

  2. 12. f(x)=x+5

  3. 13. f(x)=2x

  4. 14. f(x)=4x

  5. 15. f(x)=2x+3

  6. 16. f(x)=3x1

  7. 17. f(x)=x3+2

  8. 18. f(x)=x31

  9. 19. f(x)=(x+2)3

  10. 20. f(x)=(x1)3

  11. 21. f(x)=1x

  12. 22. f(x)=2x

  13. 23. f(x)=x

  14. 24. f(x)=x3

  15. 25. f(x)=x+4x2

  16. 26. f(x)=x+5x6

  17. 27. f(x)=2x+1x3

  18. 28. f(x)=2x3x+1

Which graphs in Exercises 2934 represent functions that have inverse functions?

  1. 29.

    A graph plots a semicircle that starts from the second quadrant and ends at the first quadrant.
  2. 30.

    A graph plots a curve that concaves down increases in the third quadrant passes through the origin and concave up increases in the first quadrant.
  3. 31.

    A graph plots a curve that falls through the negative x axis to the local minimum in the third quadrant. The curve then rises through the origin to the local maximum in the first quadrant and falls to the fourth quadrant through the positive x axis.
  4. 32.

    A graph is a line that rises from a point in the negative x axis and passes through the second quadrant and the first quadrant. After a certain point in the first quadrant, it runs parallel to the x axis.

  5. 33.

    A graph plots a concave up an increasing curve that starts from the second quadrant, passes through positive y intercept, and extends in the first quadrant.
  6. 34.

    A graph plots a concave down increasing curve starts that from the fourth quadrant, passes through positive x intercept, and extends in the first quadrant.

In Exercises 3538, use the graph of f to draw the graph of its inverse function.

  1. 35.

    A graph plots a line that rises through (0, negative 4), (2, 0), (3, 2), and (4, 4).
  2. 36.

    A graph plots a line that starts from (negative 3, 0), passes through (negative 1, 2), and (3, 3).
  3. 37.

    A graph plots a concave up an increasing curve that passes through (negative 4, 0), (0, 1), (1, 2), and (2, 4).
  4. 38.

    A graph plots a concave up decreasing curve that passes through (negative 2, 4), (negative 1, 2), (0, 1), and (4, 0).

In Exercises 3952,

  1. Find an equation for f1(x).

  2. Graph f and f1 in the same rectangular coordinate system.

  3. Use interval notation to give the domain and the range of f and f1.

  1. 39. f(x)=2x1

  2. 40. f(x)=2x3

  3. 41. f(x)=x24, x0

  4. 42. f(x)=x21, x0

  5. 43. f(x)=(x1)2, x1

  6. 44. f(x)=(x1)2, x1

  7. 45. f(x)=x31

  8. 46. f(x)=x3+1

  9. 47. f(x)=(x+2)3

  10. 48. f(x)=(x2)3

(Hint for Exercises 4952: To solve for a variable involving an nth root, raise both sides of the equation to the nth power: (yn)n=y.)

  1. 49. f(x)=x1

  2. 50. f(x)=x+2

  3. 51. f(x)=x3+1

  4. 52. f(x)=x13

Practice PLUS

In Exercises 5358, f and g are defined by the following tables. Use the tables to evaluate each composite function.

x f(x)
1 1
0 4
1 5
2 1
x g(x)
1 0
1 1
4 2
10 1
  1. 53. f(g(1))

  2. 54. f(g(4))

  3. 55. (gf)(1)

  4. 56. (gf)(0)

  5. 57. f1(g(10))

  6. 58. f1(g(1))

In Exercises 5964, let

f(x)=2x5g(x)=4x1h(x)=x2+x+2.

Evaluate the indicated function without finding an equation for the function.

  1. 59. (fg)(0)

  2. 60. (gf)(0)

  3. 61. f1(1)

  4. 62. g1(7)

  5. 63. g(f[h(1)])

  6. 64. f(g[h(1)])

Application Exercises

Way to Go Holland was the first country to establish an official bicycle policy. It currently has over 12,000 miles of paths and lanes exclusively for bicycles. The graph shows the percentage of travel by bike and by car in Holland, as well as in four other selected countries. Use the information in the graph to solve Exercises 6566.

The image shows a bar graph titled, modes of travel in selected countries by bike and by car.

Source: EUROSTAT

  1. 65.

    1. Consider a function, f, whose domain is the set of the five countries shown in the graph. Let the range be the percentage of travel by bike in each of the respective countries. Write the function f as a set of ordered pairs.

    2. Write the relation that is the inverse of f as a set of ordered pairs. Is this relation a function? Explain your answer.

  2. 66.

    1. Consider a function, f, whose domain is the set of the five countries shown in the graph. Let the range be the percentage of travel by car in each of the respective countries. Write the function f as a set of ordered pairs.

    2. Write the relation that is the inverse of f as a set of ordered pairs. Is this relation a function? Explain your answer.

  3. 67. The graph represents the probability of two people in the same room sharing a birthday as a function of the number of people in the room. Call the function f.

    A graph plots probability that two people share the same birthday versus the number of persons. The horizontal axis ranges from 0 to 60 in increments of 10, and the vertical axis ranges from 0.1 to 1.0 in increments of 0.1.
    1. Explain why f has an inverse that is a function.

    2. Describe in practical terms the meaning of f1(0.25),f1(0.5), and f1(0.7).

  4. 68. A study of 900 working women in Texas showed that their feelings changed throughout the day. As the graph indicates, the women felt better as time passed, except for a blip (that’s slang for relative maximum) at lunchtime.

    The image shows a line graph titled, average level of happiness at different times a day plots the average level of happiness versus time of day.

    Source: D. Kahneman et al., “A Survey Method for Characterizing Daily Life Experience,” Science

    1. Does the graph have an inverse that is a function? Explain your answer.

    2. Identify two or more times of day when the average happiness level is 3. Express your answers as ordered pairs.

    3. Do the ordered pairs in part (b) indicate that the graph represents a one-to-one function? Explain your answer.

  5. 69. The formula

    y=f(x)=95 x+32

    is used to convert from x degrees Celsius to y degrees Fahrenheit. The formula

    y=g(x)=59 (x32)

    is used to convert from x degrees Fahrenheit to y degrees Celsius. Show that f and g are inverse functions.

Explaining the Concepts

  1. 70. Explain how to determine if two functions are inverses of each other.

  2. 71. Describe how to find the inverse of a one-to-one function.

  3. 72. What is the horizontal line test and what does it indicate?

  4. 73. Describe how to use the graph of a one-to-one function to draw the graph of its inverse function.

  5. 74. How can a graphing utility be used to visually determine if two functions are inverses of each other?

  6. 75. What explanations can you offer for the trends shown by the graph in Exercise 68?

Technology Exercises

In Exercises 7683, use a graphing utility to graph the function. Use the graph to determine whether the function has an inverse that is a function (that is, whether the function is one-to-one).

  1. 76. f(x)=x21

  2. 77. f(x)=2x3

  3. 78. f(x)=x32

  4. 79. f(x)=x44

  5. 80. f(x)=int(x2)

  6. 81. f(x)=|x2|

  7. 82. f(x)=(x1)3

  8. 83. f(x)=16x2

In Exercises 8486, use a graphing utility to graph f and g in the same [8, 8, 1] by [5, 5, 1] viewing rectangle. In addition, graph the line y=x and visually determine if f and g are inverses.

  1. 84. f(x)=4x+4, g(x)=0.25x1

  2. 85. f(x)=1x+2, g(x)=1x2

  3. 86. f(x)=x32, g(x)=(x+2)3

Critical Thinking Exercises

Make Sense? In Exercises 8790, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 87. I found the inverse of f(x)=5x4 in my head: The reverse of multiplying by 5 and subtracting 4 is adding 4 and dividing by 5, so f1(x)=x+45.

  2. 88. I’m working with the linear function f(x)=3x+5 and I do not need to find f1 in order to determine the value of (ff1)(17).

  3. 89. When finding the inverse of a function, I interchange x and y, which reverses the domain and range between the function and its inverse.

  4. 90. I used vertical lines to determine if my graph represents a one-to-one function.

In Exercises 9194, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 91. The inverse of {(1, 4), (2, 7)} is {(2, 7), (1, 4)}.

  2. 92. The function f(x)=5 is one-to-one.

  3. 93. If f(x)=3x, then f1(x)=13x.

  4. 94. The domain of f is the same as the range of f1.

  5. 95. If f(x)=3x and g(x)=x+5, find (fg)1(x) and (g1f1)(x).

  6. 96. Show that

    f(x)=3x25x3

    is its own inverse.

  7. 97. Freedom 7 was the spacecraft that carried the first American into space in 1961. Total flight time was 15 minutes and the spacecraft reached a maximum height of 116 miles. Consider a function, s, that expresses Freedom 7’s height, s(t), in miles, after t minutes. Is s a one-to-one function? Explain your answer.

  8. 98. If f(2)=6, and f is one-to-one, find x satisfying 8+f1(x1)=10.

Group Exercise

  1. 99. In Tom Stoppard’s play Arcadia, the characters dream and talk about mathematics, including ideas involving graphing, composite functions, symmetry, and lack of symmetry in things that are tangled, mysterious, and unpredictable. Group members should read the play. Present a report on the ideas discussed by the characters that are related to concepts that we studied in this chapter. Bring in a copy of the play and read appropriate excerpts.

Retaining the Concepts

  1. 100. Solve by completing the square:

    2x25x+1=0.

    (Section P.7, Example 9)

  2. 101. The size of a television screen refers to the length of its diagonal. If the length of an HDTV screen is 28 inches and its width is 15.7 inches, what is the size of the screen to the nearest inch? (Section P.8, Example 8)

  3. 102. Solve and graph the solution set on a number line:

    3|2x1|21.

    (Section P.9, Example 9)

Preview Exercises

Exercises 103105 will help you prepare for the material covered in the next section.

  1. 103. Let (x1, y1)=(7, 2) and (x2, y2)=(1, 1). Find (x2x1)2+(y2y1)2. Express the answer in simplified radical form.

  2. 104. Use a rectangular coordinate system to graph the circle with center (1, 1) and radius 1.

  3. 105. Solve by completing the square: y26y4=0.

Objective 1: Find the distance between two points

The Distance Formula

  1. Objective 1 Find the distance between two points.

Watch Video

Using the Pythagorean Theorem, we can find the distance between the two points P1(x1, y1) and P2(x2, y2) in the rectangular coordinate system. The two points are illustrated in Figure 1.77.

Figure 1.77

A graph plots three points P sub 1 (x sub 1, y sub 1) in the third quadrant, P sub 2 (x sub 2, y sub 2) in the first quadrant, and (x sub 2, y sub 1) in the fourth quadrant.
Figure 1.77 Full Alternative Text

The distance that we need to find is represented by d and shown in blue. Notice that the distance between the two points on the dashed horizontal line is the absolute value of the difference between the x-coordinates of the points. This distance, |x2x1|, is shown in pink. Similarly, the distance between the two points on the dashed vertical line is the absolute value of the difference between the y-coordinates of the points. This distance, |y2y1|, is also shown in pink.

Because the dashed lines are horizontal and vertical, a right triangle is formed. Thus, we can use the Pythagorean Theorem to find the distance d. Squaring the lengths of the triangle’s sides results in positive numbers, so absolute value notation is not necessary.

d2=(x2x1)2+(y2y1)2Apply the Pythagorean Theorem to the righttriangle in Figure 1.77.d=±(x2x1)2+(y2y1)2Apply the square root property.d=(x2x1)2+(y2y1)2Because distance is nonnegative, write only theprincipal square root.

This result is called the distance formula.

The Distance Formula

The distance, d, between the points (x1, y1) and (x2, y2) in the rectangular coordinate system is

d=(x2x1)2+(y2y1)2.

To compute the distance between two points, find the square of the difference between the x-coordinates plus the square of the difference between the y-coordinates. The principal square root of this sum is the distance.

When using the distance formula, it does not matter which point you call (x1, y1) and which you call (x2, y2).

Example 1 Using the Distance Formula

Find the distance between (1, 4) and (3, 2). Express the answer in simplified radical form and then round to two decimal places.

Solution

We will let (x1, y1)=(1, 4) and (x2, y2)=(3, 2).

The image shows a mathematical expression to find the distance between (negative 1, 4) and (3, negative 2).

The distance between the given points is 213 units, or approximately 7.21 units. The situation is illustrated in Figure 1.78.

Figure 1.78 Finding the distance between two points

A graph plots a line that falls from (negative 1, 4) to (3, negative 2). The distance of the line is 2 radical 13 units.

Check Point 1

  • Find the distance between (1, 3) and (2, 3). Express the answer in simplified radical form and then round to two decimal places.

Objective 1: Find the distance between two points

The Distance Formula

  1. Objective 1 Find the distance between two points.

Watch Video

Using the Pythagorean Theorem, we can find the distance between the two points P1(x1, y1) and P2(x2, y2) in the rectangular coordinate system. The two points are illustrated in Figure 1.77.

Figure 1.77

A graph plots three points P sub 1 (x sub 1, y sub 1) in the third quadrant, P sub 2 (x sub 2, y sub 2) in the first quadrant, and (x sub 2, y sub 1) in the fourth quadrant.
Figure 1.77 Full Alternative Text

The distance that we need to find is represented by d and shown in blue. Notice that the distance between the two points on the dashed horizontal line is the absolute value of the difference between the x-coordinates of the points. This distance, |x2x1|, is shown in pink. Similarly, the distance between the two points on the dashed vertical line is the absolute value of the difference between the y-coordinates of the points. This distance, |y2y1|, is also shown in pink.

Because the dashed lines are horizontal and vertical, a right triangle is formed. Thus, we can use the Pythagorean Theorem to find the distance d. Squaring the lengths of the triangle’s sides results in positive numbers, so absolute value notation is not necessary.

d2=(x2x1)2+(y2y1)2Apply the Pythagorean Theorem to the righttriangle in Figure 1.77.d=±(x2x1)2+(y2y1)2Apply the square root property.d=(x2x1)2+(y2y1)2Because distance is nonnegative, write only theprincipal square root.

This result is called the distance formula.

The Distance Formula

The distance, d, between the points (x1, y1) and (x2, y2) in the rectangular coordinate system is

d=(x2x1)2+(y2y1)2.

To compute the distance between two points, find the square of the difference between the x-coordinates plus the square of the difference between the y-coordinates. The principal square root of this sum is the distance.

When using the distance formula, it does not matter which point you call (x1, y1) and which you call (x2, y2).

Example 1 Using the Distance Formula

Find the distance between (1, 4) and (3, 2). Express the answer in simplified radical form and then round to two decimal places.

Solution

We will let (x1, y1)=(1, 4) and (x2, y2)=(3, 2).

The image shows a mathematical expression to find the distance between (negative 1, 4) and (3, negative 2).

The distance between the given points is 213 units, or approximately 7.21 units. The situation is illustrated in Figure 1.78.

Figure 1.78 Finding the distance between two points

A graph plots a line that falls from (negative 1, 4) to (3, negative 2). The distance of the line is 2 radical 13 units.

Check Point 1

  • Find the distance between (1, 3) and (2, 3). Express the answer in simplified radical form and then round to two decimal places.

Objective 2: Find the midpoint of a line segment

The Midpoint Formula

  1. Objective 2 Find the midpoint of a line segment.

Watch Video

The distance formula can be used to derive a formula for finding the midpoint of a line segment between two given points. The formula is given as follows:

The Midpoint Formula

Consider a line segment whose endpoints are (x1, y1) and (x2, y2). The coordinates of the segment’s midpoint are

(x1+x22, y1+y22).

To find the midpoint, take the average of the two x-coordinates and the average of the two y-coordinates.

Example 2 Using the Midpoint Formula

Find the midpoint of the line segment with endpoints (1, 6) and (8, 4).

Solution

To find the coordinates of the midpoint, we average the coordinates of the endpoints.

The image shows a solution of a mathematical expression using the midpoint formula.

Figure 1.79 illustrates that the point ( 72, 5) is midway between the points (1, 6) and (8, 4).

Figure 1.79 Finding a line segment’s midpoint

A graph plots a line segment that starts from (negative 8, negative 4) passes through the midpoint (negative 7 halves, negative 5), and ends at (1, negative 6).

Check Point 2

  • Find the midpoint of the line segment with endpoints (1, 2) and (7, 3).

Circles

Our goal is to translate a circle’s geometric definition into an equation. We begin with this geometric definition.

Definition of a Circle

A circle is the set of all points in a plane that are equidistant from a fixed point, called the center. The fixed distance from the circle’s center to any point on the circle is called the radius.

Figure 1.80 is our starting point for obtaining a circle’s equation. We’ve placed the circle into a rectangular coordinate system. The circle’s center is (h, k) and its radius is r. We let (x, y) represent the coordinates of any point on the circle.

Figure 1.80 A circle centered at (h, k) with radius r

A graph plots a circle in the first quadrant centered at (h, k) with radius r. The circumference of the circle passes through the point (x, y).

What does the geometric definition of a circle tell us about the point (x, y) in Figure 1.80? The point is on the circle if and only if its distance from the center is r. We can use the distance formula to express this idea algebraically:

The square root of start expression left parenthesis x minus h right parenthesis squared + left parenthesis y minus k right parenthesis squared end expression = r.

Squaring both sides of (xh)2+(yk)2=r yields the standard form of the equation of a circle.

Objective 3: Write the standard form of a circle’s equation

  1. Objective 3 Write the standard form of a circle’s equation.

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The Standard Form of the Equation of a Circle

The standard form of the equation of a circle with center (h, k) and radius r is

(xh)2+(yk)2=r2.

Example 3 Finding the Standard Form of a Circle’s Equation

Write the standard form of the equation of the circle with center (0, 0) and radius 2. Graph the circle.

Solution

The center is (0, 0). Because the center is represented as (h, k) in the standard form of the equation, h=0 and k=0. The radius is 2, so we will let r=2 in the equation.

(xh)2+(yk)2=r2This is the standard form of a circle's equation.(x0)2+(y0)2=22Substitute 0 for h,0 for k,and 2 for r.x2+y2=4Simplify.

The standard form of the equation of the circle is x2+y2=4. Figure 1.81 shows the graph.

Figure 1.81 The graph of x2+y2=4

The graph of x squared + y squared = 4 is a circle passing through all quadrants centered at the origin with a radius of 2 units. The circumference of the circle passes through the plotted point (x, y).

Check Point 3

  • Write the standard form of the equation of the circle with center (0, 0) and radius 4.

Example 3 and Check Point 3 involved circles centered at the origin. The standard form of the equation of all such circles is x2+y2=r2, where r is the circle’s radius. Now, let’s consider a circle whose center is not at the origin.

Example 4 Finding the Standard Form of a Circle’s Equation

Write the standard form of the equation of the circle with center (2, 3) and radius 4.

Solution

The center is (2, 3). Because the center is represented as (h, k) in the standard form of the equation, h=2 and k=3. The radius is 4, so we will let r=4 in the equation.

(xh)2+(yk)2=r2This is the standard form of a circle's equation.[x(2)]2+(y3)2=42Substitute 2 for h, 3 for k, and 4 for r.(x+2)2+(y3)2=16Simplify.

The standard form of the equation of the circle is (x+2)2+(y3)2=16.

Check Point 4

  • Write the standard form of the equation of the circle with center (0, 6) and radius 10.

Objective 4: Give the center and radius of a circle whose equation is in standard form

  1. Objective 4 Give the center and radius of a circle whose equation is in standard form.

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Example 5 Using the Standard Form of a Circle’s Equation to Graph the Circle

  1. Find the center and radius of the circle whose equation is

    (x2)2+(y+4)2=9.
  2. Graph the equation.

  3. Use the graph to identify the relation’s domain and range.

Solution

  1. We begin by finding the circle’s center, (h, k), and its radius, r. We can find the values for h, k, and r by comparing the given equation to the standard form of the equation of a circle, (xh)2+(yk)2=r2.

    The image shows a mathematical expression to solve the equation of a circle.

    We see that h=2, k=4, and r=3. Thus, the circle has center (h, k)=(2, 4) and a radius of 3 units.

  2. To graph this circle, first locate the center (2, 4). Because the radius is 3, you can locate at least four points on the circle by going out 3 units to the right, to the left, up, and down from the center.

    The points 3 units to the right and to the left of (2, 4) are (5, 4) and (1, 4), respectively. The points 3 units up and down from (2, 4) are (2, 1) and (2, 7), respectively.

    Using these points, we obtain the graph in Figure 1.82.

    Figure 1.82 The graph of (x2)2+(y+4)2=9

    The graph is a circle in the third and fourth quadrant, centered at (2, negative 4) with a radius of 3 units passes through the points (5, negative 4), (2, negative 1), (negative 1, negative 4), and (2, negative 7).
  3. The four points that we located on the circle can be used to determine the relation’s domain and range. The points (1, 4) and (5, 4) show that values of x extend from 1 to 5, inclusive:

    Domain=[1, 5].

    The points (2, 7) and (2, 1) show that values of y extend from 7 to 1, inclusive:

    Range=[7, 1].

Check Point 5

    1. Find the center and radius of the circle whose equation is

      (x+3)2+(y1)2=4.
    2. Graph the equation.

    3. Use the graph to identify the relation’s domain and range.

If we square x2 and y+4 in the standard form of the equation in Example 5, we obtain another form for the circle’s equation.

(x2)2+(y+4)2=9This is the standard form of the equation inExample 5.x24x+4+y2+8y+16=9Square x2 and y+4.x2+y24x+8y+20=9Combine constants and rearrange terms.x2+y24x+8y+11=0Subtract 9 from both sides.

This result, x2+y24x+8y+11=0, suggests that an equation in the form x2+y2+Dx+Ey+F=0 can represent a circle. This is called the general form of the equation of a circle.

The General Form of the Equation of a Circle

The general form of the equation of a circle is

x2+y2+Dx+Ey+F=0,

where D, E, and F are real numbers.

Objective 4: Give the center and radius of a circle whose equation is in standard form

  1. Objective 4 Give the center and radius of a circle whose equation is in standard form.

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Example 5 Using the Standard Form of a Circle’s Equation to Graph the Circle

  1. Find the center and radius of the circle whose equation is

    (x2)2+(y+4)2=9.
  2. Graph the equation.

  3. Use the graph to identify the relation’s domain and range.

Solution

  1. We begin by finding the circle’s center, (h, k), and its radius, r. We can find the values for h, k, and r by comparing the given equation to the standard form of the equation of a circle, (xh)2+(yk)2=r2.

    The image shows a mathematical expression to solve the equation of a circle.

    We see that h=2, k=4, and r=3. Thus, the circle has center (h, k)=(2, 4) and a radius of 3 units.

  2. To graph this circle, first locate the center (2, 4). Because the radius is 3, you can locate at least four points on the circle by going out 3 units to the right, to the left, up, and down from the center.

    The points 3 units to the right and to the left of (2, 4) are (5, 4) and (1, 4), respectively. The points 3 units up and down from (2, 4) are (2, 1) and (2, 7), respectively.

    Using these points, we obtain the graph in Figure 1.82.

    Figure 1.82 The graph of (x2)2+(y+4)2=9

    The graph is a circle in the third and fourth quadrant, centered at (2, negative 4) with a radius of 3 units passes through the points (5, negative 4), (2, negative 1), (negative 1, negative 4), and (2, negative 7).
  3. The four points that we located on the circle can be used to determine the relation’s domain and range. The points (1, 4) and (5, 4) show that values of x extend from 1 to 5, inclusive:

    Domain=[1, 5].

    The points (2, 7) and (2, 1) show that values of y extend from 7 to 1, inclusive:

    Range=[7, 1].

Check Point 5

    1. Find the center and radius of the circle whose equation is

      (x+3)2+(y1)2=4.
    2. Graph the equation.

    3. Use the graph to identify the relation’s domain and range.

If we square x2 and y+4 in the standard form of the equation in Example 5, we obtain another form for the circle’s equation.

(x2)2+(y+4)2=9This is the standard form of the equation inExample 5.x24x+4+y2+8y+16=9Square x2 and y+4.x2+y24x+8y+20=9Combine constants and rearrange terms.x2+y24x+8y+11=0Subtract 9 from both sides.

This result, x2+y24x+8y+11=0, suggests that an equation in the form x2+y2+Dx+Ey+F=0 can represent a circle. This is called the general form of the equation of a circle.

The General Form of the Equation of a Circle

The general form of the equation of a circle is

x2+y2+Dx+Ey+F=0,

where D, E, and F are real numbers.

Objective 5: Convert the general form of a circle’s equation to standard form

  1. Objective 5 Convert the general form of a circle’s equation to standard form.

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We can convert the general form of the equation of a circle to the standard form (xh)2+(yk)2=r2. We do so by completing the square on x and y. Let’s see how this is done.

Example 6 Converting the General Form of a Circle’s Equation to Standard Form and Graphing the Circle

Write in standard form and graph: x2+y2+4x6y23=0.

Solution

Because we plan to complete the square on both x and y, let’s rearrange the terms so that x-terms are arranged in descending order, y-terms are arranged in descending order, and the constant term appears on the right.

The image shows a mathematical expression to solve the equation x squared + y squared + 4 x minus 6 y minus 23 = 0.

This last equation, (x+2)2+(y3)2=36, is in standard form. We can identify the circle’s center and radius by comparing this equation to the standard form of the equation of a circle, (xh)2+(yk)2=r2.

The image shows a mathematical expression to find the radius of a circle.

We use the center, (h, k)=(2, 3), and the radius, r=6, to graph the circle. The graph is shown in Figure 1.83.

Figure 1.83 The graph of (x+2)2+(y3)2=36

A graph is a circle in all quadrants, centered at (negative 2, 3) with a radius of 6 units.

Check Point 6

  • Write in standard form and graph:

    x2+y2+4x4y1=0.
1.9: Exercise Set

1.9 Exercise Set

Practice Exercises

In Exercises 118, find the distance between each pair of points. If necessary, express answers in simplified radical form and then round to two decimal places.

  1. 1. (2, 3) and (14, 8)

  2. 2. (5, 1) and (8, 5)

  3. 3. (4, 1) and (6, 3)

  4. 4. (2, 3) and (1, 5)

  5. 5. (0, 0) and (3, 4)

  6. 6. (0, 0) and (3, 4)

  7. 7. (2, 6) and (3, 4)

  8. 8. (4, 1) and (2, 3)

  9. 9. (0, 3) and (4, 1)

  10. 10. (0, 2) and (4, 3)

  11. 11. (3.5, 8.2) and (0.5, 6.2)

  12. 12. (2.6, 1.3) and (1.6, 5.7)

  13. 13. (0, 3) and (5, 0)

  14. 14. (0, 2) and (7, 0)

  15. 15. (33, 5) and (3, 45)

  16. 16. (23, 6) and (3, 56)

  17. 17. (73, 15) and (13, 65)

  18. 18. ( 14,  17) and (34, 67)

In Exercises 1930, find the midpoint of each line segment with the given endpoints.

  1. 19. (6, 8) and (2, 4)

  2. 20. (10, 4) and (2, 6)

  3. 21. (2, 8) and (6, 2)

  4. 22. (4, 7) and (1, 3)

  5. 23. (3, 4) and (6, 8)

  6. 24. (2, 1) and (8, 6)

  7. 25. ( 72, 32) and ( 52,  112)

  8. 26. ( 25, 715) and ( 25,  415)

  9. 27. (8, 35) and (6, 75)

  10. 28. (73, 6) and (33, 2)

  11. 29. (18, 4) and (2, 4)

  12. 30. (50, 6) and (2, 6)

In Exercises 3140, write the standard form of the equation of the circle with the given center and radius.

  1. 31. Center (0, 0), r=7

  2. 32. Center (0, 0), r=8

  3. 33. Center (3, 2), r=5

  4. 34. Center (2, 1), r=4

  5. 35. Center (1, 4), r=2

  6. 36. Center (3, 5), r=3

  7. 37. Center (3, 1), r=3

  8. 38. Center (5, 3), r=5

  9. 39. Center (4, 0), r=10

  10. 40. Center (2, 0), r=6

In Exercises 4152, give the center and radius of the circle described by the equation and graph each equation. Use the graph to identify the relation’s domain and range.

  1. 41. x2+y2=16

  2. 42. x2+y2=49

  3. 43. (x3)2+(y1)2=36

  4. 44. (x2)2+(y3)2=16

  5. 45. (x+3)2+(y2)2=4

  6. 46. (x+1)2+(y4)2=25

  7. 47. (x+2)2+(y+2)2=4

  8. 48. (x+4)2+(y+5)2=36

  9. 49. x2+(y1)2=1

  10. 50. x2+(y2)2=4

  11. 51. (x+1)2+y2=25

  12. 52. (x+2)2+y2=16

In Exercises 5364, complete the square and write the equation in standard form. Then give the center and radius of each circle and graph the equation.

  1. 53. x2+y2+6x+2y+6=0

  2. 54. x2+y2+8x+4y+16=0

  3. 55. x2+y210x6y30=0

  4. 56. x2+y24x12y9=0

  5. 57. x2+y2+8x2y8=0

  6. 58. x2+y2+12x6y4=0

  7. 59. x22x+y215=0

  8. 60. x2+y26y7=0

  9. 61. x2+y2x+2y+1=0

  10. 62. x2+y2+x+y12=0

  11. 63. x2+y2+3x2y1=0

  12. 64. x2+y2+3x+5y+94=0

Practice PLUS

In Exercises 6566, a line segment through the center of each circle intersects the circle at the points shown.

  1. Find the coordinates of the circle’s center.

  2. Find the radius of the circle.

  3. Use your answers from parts (a) and (b) to write the standard form of the circle’s equation.

  1. 65.

    A graph displays a circle in the first quadrant. The line passes through the plotted points (7, 11) and (3, 9). A line segment connects the plotted points.
  2. 66.

    A graph displays a circle in the first quadrant. The circle passes through the plotted points (3, 6) and (5, 4). A line segment connects the plotted points.

In Exercises 6770, graph both equations in the same rectangular coordinate system and find all points of intersection. Then show that these ordered pairs satisfy the equations.

  1. 67. x2+y2=16xy=4

  2. 68. x2+y2=9xy=3

  3. 69. (x2)2+(y+3)2=4y=x3

  4. 70. (x3)2+(y+1)2=9y=x1

Application Exercises

The smartphone screen shows coordinates of six cities from a rectangular coordinate system placed on North America by long-distance telephone companies. Each unit in this system represents 0.1 mile.

Source: Peter H. Dana

In Exercises 7172, use this information to find the distance, to the nearest mile, between each pair of cities.

  1. 71. Boston and San Francisco

  2. 72. New Orleans and Houston

  3. 73. A rectangular coordinate system with coordinates in miles is placed with the origin at the center of Los Angeles. The figure indicates that the University of Southern California is located 2.4 miles west and 2.7 miles south of central Los Angeles. A seismograph on the campus shows that a small earthquake occurred. The quake’s epicenter is estimated to be approximately 30 miles from the university. Write the standard form of the equation for the set of points that could be the epicenter of the quake.

    A rectangular coordinate system depicts a Los Angeles city located at the origin (0, 0) and University of Southern California located at (negative 2.4, negative 2. 7).
  4. 74. The Ferris wheel in the figure has a radius of 68 feet. The clearance between the wheel and the ground is 14 feet. The rectangular coordinate system shown has its origin on the ground directly below the center of the wheel. Use the coordinate system to write the equation of the circular wheel.

    A coordinated system displays a Ferris wheel with a radius of 68 feet, and the bottom of the wheel passes 14 feet above the ground.

Explaining the Concepts

  1. 75. In your own words, describe how to find the distance between two points in the rectangular coordinate system.

  2. 76. In your own words, describe how to find the midpoint of a line segment if its endpoints are known.

  3. 77. What is a circle? Without using variables, describe how the definition of a circle can be used to obtain a form of its equation.

  4. 78. Give an example of a circle’s equation in standard form. Describe how to find the center and radius for this circle.

  5. 79. How is the standard form of a circle’s equation obtained from its general form?

  6. 80. Does (x3)2+(y5)2=0 represent the equation of a circle? If not, describe the graph of this equation.

  7. 81. Does (x3)2+(y5)2=25 represent the equation of a circle? What sort of set is the graph of this equation?

  8. 82. Write and solve a problem about the flying time between a pair of cities shown on the smartphone screen for Exercises 7172. Do not use the pairs in Exercise 71 or Exercise 72. Begin by determining a reasonable average speed, in miles per hour, for a jet flying between the cities.

Technology Exercises

In Exercises 8385, use a graphing utility to graph each circle whose equation is given. Use a square setting for the viewing window.

  1. 83. x2+y2=25

  2. 84. (y+1)2=36(x3)2

  3. 85. x2+10x+y24y20=0

Critical Thinking Exercises

Make Sense? In Exercises 8689, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 86. I’ve noticed that in mathematics, one topic often leads logically to a new topic:

    Pythagorean Theorem leads to distance formula leads to an equation of circles.
  2. 87. To avoid sign errors when finding h and k, I place parentheses around the numbers that follow the subtraction signs in a circle’s equation.

  3. 88. I used the equation (x+1)2+(y5)2=4 to identify the circle’s center and radius.

  4. 89. My graph of (x2)2+(y+1)2=16 is my graph of x2+y2=16 translated 2 units right and 1 unit down.

In Exercises 9093, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 90. The equation of the circle whose center is at the origin with radius 16 is x2+y2=16.

  2. 91. The graph of (x3)2+(y+5)2=36 is a circle with radius 6 centered at (3, 5).

  3. 92. The graph of (x4)+(y+6)=25 is a circle with radius 5 centered at (4, 6).

  4. 93. The graph of (x3)2+(y+5)2=36 is a circle with radius 6 centered at (3, 5).

  5. 94. Show that the points A(1, 1+d), B(3, 3+d), and C(6, 6+d) are collinear (lie along a straight line) by showing that the distance from A to B plus the distance from B to C equals the distance from A to C.

  6. 95. Prove the midpoint formula by using the following procedure.

    1. Show that the distance between (x1, y1) and (x1+x22, y1+y22) is equal to the distance between (x2, y2) and (x1+x22, y1+y22).

    2. Use the procedure from Exercise 94 and the distances from part (a) to show that the points (x1, y1), (x1+x22, y1+y22), and (x2, y2) are collinear.

  7. 96. Find the area of the donut-shaped region bounded by the graphs of (x2)2+(y+3)2=25 and (x2)2+(y+3)2=36.

  8. 97. A tangent line to a circle is a line that intersects the circle at exactly one point. The tangent line is perpendicular to the radius of the circle at this point of contact. Write an equation in point-slope form for the line tangent to the circle whose equation is x2+y2=25 at the point (3, 4).

Retaining the Concepts

  1. 98. Determine whether the graph of x2y3=2 is symmetric with respect to the y-axis, the x-axis, the origin, more than one of these, or none of these.

    (Section 1.3, Examples 2 and 3)

  2. 99. Determine whether each relation is a function. Give the domain and range for each relation.

    1. {(1, 6), (1, 7), (1, 8)}

    2. {(6, 1), (7, 1), (8, 1)}

    (Section 1.2, Example 2)

  3. 100. Solve: 2x+34x+5=6x2+8x+15.

    (Section P.7, Example 3)

Preview Exercises

Exercises 101103 will help you prepare for the material covered in the next section.

  1. 101. Write an algebraic expression for the fare increase if a $200 plane ticket is increased to x dollars.

  2. 102. Find the perimeter and the area of each rectangle with the given dimensions:

    1. 40 yards by 30 yards

    2. 50 yards by 20 yards.

  3. 103. Solve for h: πr2h=22. Then rewrite 2πr2+2πrh in terms of r.

1.9: Exercise Set

1.9 Exercise Set

Practice Exercises

In Exercises 118, find the distance between each pair of points. If necessary, express answers in simplified radical form and then round to two decimal places.

  1. 1. (2, 3) and (14, 8)

  2. 2. (5, 1) and (8, 5)

  3. 3. (4, 1) and (6, 3)

  4. 4. (2, 3) and (1, 5)

  5. 5. (0, 0) and (3, 4)

  6. 6. (0, 0) and (3, 4)

  7. 7. (2, 6) and (3, 4)

  8. 8. (4, 1) and (2, 3)

  9. 9. (0, 3) and (4, 1)

  10. 10. (0, 2) and (4, 3)

  11. 11. (3.5, 8.2) and (0.5, 6.2)

  12. 12. (2.6, 1.3) and (1.6, 5.7)

  13. 13. (0, 3) and (5, 0)

  14. 14. (0, 2) and (7, 0)

  15. 15. (33, 5) and (3, 45)

  16. 16. (23, 6) and (3, 56)

  17. 17. (73, 15) and (13, 65)

  18. 18. ( 14,  17) and (34, 67)

In Exercises 1930, find the midpoint of each line segment with the given endpoints.

  1. 19. (6, 8) and (2, 4)

  2. 20. (10, 4) and (2, 6)

  3. 21. (2, 8) and (6, 2)

  4. 22. (4, 7) and (1, 3)

  5. 23. (3, 4) and (6, 8)

  6. 24. (2, 1) and (8, 6)

  7. 25. ( 72, 32) and ( 52,  112)

  8. 26. ( 25, 715) and ( 25,  415)

  9. 27. (8, 35) and (6, 75)

  10. 28. (73, 6) and (33, 2)

  11. 29. (18, 4) and (2, 4)

  12. 30. (50, 6) and (2, 6)

In Exercises 3140, write the standard form of the equation of the circle with the given center and radius.

  1. 31. Center (0, 0), r=7

  2. 32. Center (0, 0), r=8

  3. 33. Center (3, 2), r=5

  4. 34. Center (2, 1), r=4

  5. 35. Center (1, 4), r=2

  6. 36. Center (3, 5), r=3

  7. 37. Center (3, 1), r=3

  8. 38. Center (5, 3), r=5

  9. 39. Center (4, 0), r=10

  10. 40. Center (2, 0), r=6

In Exercises 4152, give the center and radius of the circle described by the equation and graph each equation. Use the graph to identify the relation’s domain and range.

  1. 41. x2+y2=16

  2. 42. x2+y2=49

  3. 43. (x3)2+(y1)2=36

  4. 44. (x2)2+(y3)2=16

  5. 45. (x+3)2+(y2)2=4

  6. 46. (x+1)2+(y4)2=25

  7. 47. (x+2)2+(y+2)2=4

  8. 48. (x+4)2+(y+5)2=36

  9. 49. x2+(y1)2=1

  10. 50. x2+(y2)2=4

  11. 51. (x+1)2+y2=25

  12. 52. (x+2)2+y2=16

In Exercises 5364, complete the square and write the equation in standard form. Then give the center and radius of each circle and graph the equation.

  1. 53. x2+y2+6x+2y+6=0

  2. 54. x2+y2+8x+4y+16=0

  3. 55. x2+y210x6y30=0

  4. 56. x2+y24x12y9=0

  5. 57. x2+y2+8x2y8=0

  6. 58. x2+y2+12x6y4=0

  7. 59. x22x+y215=0

  8. 60. x2+y26y7=0

  9. 61. x2+y2x+2y+1=0

  10. 62. x2+y2+x+y12=0

  11. 63. x2+y2+3x2y1=0

  12. 64. x2+y2+3x+5y+94=0

Practice PLUS

In Exercises 6566, a line segment through the center of each circle intersects the circle at the points shown.

  1. Find the coordinates of the circle’s center.

  2. Find the radius of the circle.

  3. Use your answers from parts (a) and (b) to write the standard form of the circle’s equation.

  1. 65.

    A graph displays a circle in the first quadrant. The line passes through the plotted points (7, 11) and (3, 9). A line segment connects the plotted points.
  2. 66.

    A graph displays a circle in the first quadrant. The circle passes through the plotted points (3, 6) and (5, 4). A line segment connects the plotted points.

In Exercises 6770, graph both equations in the same rectangular coordinate system and find all points of intersection. Then show that these ordered pairs satisfy the equations.

  1. 67. x2+y2=16xy=4

  2. 68. x2+y2=9xy=3

  3. 69. (x2)2+(y+3)2=4y=x3

  4. 70. (x3)2+(y+1)2=9y=x1

Application Exercises

The smartphone screen shows coordinates of six cities from a rectangular coordinate system placed on North America by long-distance telephone companies. Each unit in this system represents 0.1 mile.

Source: Peter H. Dana

In Exercises 7172, use this information to find the distance, to the nearest mile, between each pair of cities.

  1. 71. Boston and San Francisco

  2. 72. New Orleans and Houston

  3. 73. A rectangular coordinate system with coordinates in miles is placed with the origin at the center of Los Angeles. The figure indicates that the University of Southern California is located 2.4 miles west and 2.7 miles south of central Los Angeles. A seismograph on the campus shows that a small earthquake occurred. The quake’s epicenter is estimated to be approximately 30 miles from the university. Write the standard form of the equation for the set of points that could be the epicenter of the quake.

    A rectangular coordinate system depicts a Los Angeles city located at the origin (0, 0) and University of Southern California located at (negative 2.4, negative 2. 7).
  4. 74. The Ferris wheel in the figure has a radius of 68 feet. The clearance between the wheel and the ground is 14 feet. The rectangular coordinate system shown has its origin on the ground directly below the center of the wheel. Use the coordinate system to write the equation of the circular wheel.

    A coordinated system displays a Ferris wheel with a radius of 68 feet, and the bottom of the wheel passes 14 feet above the ground.

Explaining the Concepts

  1. 75. In your own words, describe how to find the distance between two points in the rectangular coordinate system.

  2. 76. In your own words, describe how to find the midpoint of a line segment if its endpoints are known.

  3. 77. What is a circle? Without using variables, describe how the definition of a circle can be used to obtain a form of its equation.

  4. 78. Give an example of a circle’s equation in standard form. Describe how to find the center and radius for this circle.

  5. 79. How is the standard form of a circle’s equation obtained from its general form?

  6. 80. Does (x3)2+(y5)2=0 represent the equation of a circle? If not, describe the graph of this equation.

  7. 81. Does (x3)2+(y5)2=25 represent the equation of a circle? What sort of set is the graph of this equation?

  8. 82. Write and solve a problem about the flying time between a pair of cities shown on the smartphone screen for Exercises 7172. Do not use the pairs in Exercise 71 or Exercise 72. Begin by determining a reasonable average speed, in miles per hour, for a jet flying between the cities.

Technology Exercises

In Exercises 8385, use a graphing utility to graph each circle whose equation is given. Use a square setting for the viewing window.

  1. 83. x2+y2=25

  2. 84. (y+1)2=36(x3)2

  3. 85. x2+10x+y24y20=0

Critical Thinking Exercises

Make Sense? In Exercises 8689, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 86. I’ve noticed that in mathematics, one topic often leads logically to a new topic:

    Pythagorean Theorem leads to distance formula leads to an equation of circles.
  2. 87. To avoid sign errors when finding h and k, I place parentheses around the numbers that follow the subtraction signs in a circle’s equation.

  3. 88. I used the equation (x+1)2+(y5)2=4 to identify the circle’s center and radius.

  4. 89. My graph of (x2)2+(y+1)2=16 is my graph of x2+y2=16 translated 2 units right and 1 unit down.

In Exercises 9093, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 90. The equation of the circle whose center is at the origin with radius 16 is x2+y2=16.

  2. 91. The graph of (x3)2+(y+5)2=36 is a circle with radius 6 centered at (3, 5).

  3. 92. The graph of (x4)+(y+6)=25 is a circle with radius 5 centered at (4, 6).

  4. 93. The graph of (x3)2+(y+5)2=36 is a circle with radius 6 centered at (3, 5).

  5. 94. Show that the points A(1, 1+d), B(3, 3+d), and C(6, 6+d) are collinear (lie along a straight line) by showing that the distance from A to B plus the distance from B to C equals the distance from A to C.

  6. 95. Prove the midpoint formula by using the following procedure.

    1. Show that the distance between (x1, y1) and (x1+x22, y1+y22) is equal to the distance between (x2, y2) and (x1+x22, y1+y22).

    2. Use the procedure from Exercise 94 and the distances from part (a) to show that the points (x1, y1), (x1+x22, y1+y22), and (x2, y2) are collinear.

  7. 96. Find the area of the donut-shaped region bounded by the graphs of (x2)2+(y+3)2=25 and (x2)2+(y+3)2=36.

  8. 97. A tangent line to a circle is a line that intersects the circle at exactly one point. The tangent line is perpendicular to the radius of the circle at this point of contact. Write an equation in point-slope form for the line tangent to the circle whose equation is x2+y2=25 at the point (3, 4).

Retaining the Concepts

  1. 98. Determine whether the graph of x2y3=2 is symmetric with respect to the y-axis, the x-axis, the origin, more than one of these, or none of these.

    (Section 1.3, Examples 2 and 3)

  2. 99. Determine whether each relation is a function. Give the domain and range for each relation.

    1. {(1, 6), (1, 7), (1, 8)}

    2. {(6, 1), (7, 1), (8, 1)}

    (Section 1.2, Example 2)

  3. 100. Solve: 2x+34x+5=6x2+8x+15.

    (Section P.7, Example 3)

Preview Exercises

Exercises 101103 will help you prepare for the material covered in the next section.

  1. 101. Write an algebraic expression for the fare increase if a $200 plane ticket is increased to x dollars.

  2. 102. Find the perimeter and the area of each rectangle with the given dimensions:

    1. 40 yards by 30 yards

    2. 50 yards by 20 yards.

  3. 103. Solve for h: πr2h=22. Then rewrite 2πr2+2πrh in terms of r.

1.9: Exercise Set

1.9 Exercise Set

Practice Exercises

In Exercises 118, find the distance between each pair of points. If necessary, express answers in simplified radical form and then round to two decimal places.

  1. 1. (2, 3) and (14, 8)

  2. 2. (5, 1) and (8, 5)

  3. 3. (4, 1) and (6, 3)

  4. 4. (2, 3) and (1, 5)

  5. 5. (0, 0) and (3, 4)

  6. 6. (0, 0) and (3, 4)

  7. 7. (2, 6) and (3, 4)

  8. 8. (4, 1) and (2, 3)

  9. 9. (0, 3) and (4, 1)

  10. 10. (0, 2) and (4, 3)

  11. 11. (3.5, 8.2) and (0.5, 6.2)

  12. 12. (2.6, 1.3) and (1.6, 5.7)

  13. 13. (0, 3) and (5, 0)

  14. 14. (0, 2) and (7, 0)

  15. 15. (33, 5) and (3, 45)

  16. 16. (23, 6) and (3, 56)

  17. 17. (73, 15) and (13, 65)

  18. 18. ( 14,  17) and (34, 67)

In Exercises 1930, find the midpoint of each line segment with the given endpoints.

  1. 19. (6, 8) and (2, 4)

  2. 20. (10, 4) and (2, 6)

  3. 21. (2, 8) and (6, 2)

  4. 22. (4, 7) and (1, 3)

  5. 23. (3, 4) and (6, 8)

  6. 24. (2, 1) and (8, 6)

  7. 25. ( 72, 32) and ( 52,  112)

  8. 26. ( 25, 715) and ( 25,  415)

  9. 27. (8, 35) and (6, 75)

  10. 28. (73, 6) and (33, 2)

  11. 29. (18, 4) and (2, 4)

  12. 30. (50, 6) and (2, 6)

In Exercises 3140, write the standard form of the equation of the circle with the given center and radius.

  1. 31. Center (0, 0), r=7

  2. 32. Center (0, 0), r=8

  3. 33. Center (3, 2), r=5

  4. 34. Center (2, 1), r=4

  5. 35. Center (1, 4), r=2

  6. 36. Center (3, 5), r=3

  7. 37. Center (3, 1), r=3

  8. 38. Center (5, 3), r=5

  9. 39. Center (4, 0), r=10

  10. 40. Center (2, 0), r=6

In Exercises 4152, give the center and radius of the circle described by the equation and graph each equation. Use the graph to identify the relation’s domain and range.

  1. 41. x2+y2=16

  2. 42. x2+y2=49

  3. 43. (x3)2+(y1)2=36

  4. 44. (x2)2+(y3)2=16

  5. 45. (x+3)2+(y2)2=4

  6. 46. (x+1)2+(y4)2=25

  7. 47. (x+2)2+(y+2)2=4

  8. 48. (x+4)2+(y+5)2=36

  9. 49. x2+(y1)2=1

  10. 50. x2+(y2)2=4

  11. 51. (x+1)2+y2=25

  12. 52. (x+2)2+y2=16

In Exercises 5364, complete the square and write the equation in standard form. Then give the center and radius of each circle and graph the equation.

  1. 53. x2+y2+6x+2y+6=0

  2. 54. x2+y2+8x+4y+16=0

  3. 55. x2+y210x6y30=0

  4. 56. x2+y24x12y9=0

  5. 57. x2+y2+8x2y8=0

  6. 58. x2+y2+12x6y4=0

  7. 59. x22x+y215=0

  8. 60. x2+y26y7=0

  9. 61. x2+y2x+2y+1=0

  10. 62. x2+y2+x+y12=0

  11. 63. x2+y2+3x2y1=0

  12. 64. x2+y2+3x+5y+94=0

Practice PLUS

In Exercises 6566, a line segment through the center of each circle intersects the circle at the points shown.

  1. Find the coordinates of the circle’s center.

  2. Find the radius of the circle.

  3. Use your answers from parts (a) and (b) to write the standard form of the circle’s equation.

  1. 65.

    A graph displays a circle in the first quadrant. The line passes through the plotted points (7, 11) and (3, 9). A line segment connects the plotted points.
  2. 66.

    A graph displays a circle in the first quadrant. The circle passes through the plotted points (3, 6) and (5, 4). A line segment connects the plotted points.

In Exercises 6770, graph both equations in the same rectangular coordinate system and find all points of intersection. Then show that these ordered pairs satisfy the equations.

  1. 67. x2+y2=16xy=4

  2. 68. x2+y2=9xy=3

  3. 69. (x2)2+(y+3)2=4y=x3

  4. 70. (x3)2+(y+1)2=9y=x1

Application Exercises

The smartphone screen shows coordinates of six cities from a rectangular coordinate system placed on North America by long-distance telephone companies. Each unit in this system represents 0.1 mile.

Source: Peter H. Dana

In Exercises 7172, use this information to find the distance, to the nearest mile, between each pair of cities.

  1. 71. Boston and San Francisco

  2. 72. New Orleans and Houston

  3. 73. A rectangular coordinate system with coordinates in miles is placed with the origin at the center of Los Angeles. The figure indicates that the University of Southern California is located 2.4 miles west and 2.7 miles south of central Los Angeles. A seismograph on the campus shows that a small earthquake occurred. The quake’s epicenter is estimated to be approximately 30 miles from the university. Write the standard form of the equation for the set of points that could be the epicenter of the quake.

    A rectangular coordinate system depicts a Los Angeles city located at the origin (0, 0) and University of Southern California located at (negative 2.4, negative 2. 7).
  4. 74. The Ferris wheel in the figure has a radius of 68 feet. The clearance between the wheel and the ground is 14 feet. The rectangular coordinate system shown has its origin on the ground directly below the center of the wheel. Use the coordinate system to write the equation of the circular wheel.

    A coordinated system displays a Ferris wheel with a radius of 68 feet, and the bottom of the wheel passes 14 feet above the ground.

Explaining the Concepts

  1. 75. In your own words, describe how to find the distance between two points in the rectangular coordinate system.

  2. 76. In your own words, describe how to find the midpoint of a line segment if its endpoints are known.

  3. 77. What is a circle? Without using variables, describe how the definition of a circle can be used to obtain a form of its equation.

  4. 78. Give an example of a circle’s equation in standard form. Describe how to find the center and radius for this circle.

  5. 79. How is the standard form of a circle’s equation obtained from its general form?

  6. 80. Does (x3)2+(y5)2=0 represent the equation of a circle? If not, describe the graph of this equation.

  7. 81. Does (x3)2+(y5)2=25 represent the equation of a circle? What sort of set is the graph of this equation?

  8. 82. Write and solve a problem about the flying time between a pair of cities shown on the smartphone screen for Exercises 7172. Do not use the pairs in Exercise 71 or Exercise 72. Begin by determining a reasonable average speed, in miles per hour, for a jet flying between the cities.

Technology Exercises

In Exercises 8385, use a graphing utility to graph each circle whose equation is given. Use a square setting for the viewing window.

  1. 83. x2+y2=25

  2. 84. (y+1)2=36(x3)2

  3. 85. x2+10x+y24y20=0

Critical Thinking Exercises

Make Sense? In Exercises 8689, determine whether each statement makes sense or does not make sense, and explain your reasoning.

  1. 86. I’ve noticed that in mathematics, one topic often leads logically to a new topic:

    Pythagorean Theorem leads to distance formula leads to an equation of circles.
  2. 87. To avoid sign errors when finding h and k, I place parentheses around the numbers that follow the subtraction signs in a circle’s equation.

  3. 88. I used the equation (x+1)2+(y5)2=4 to identify the circle’s center and radius.

  4. 89. My graph of (x2)2+(y+1)2=16 is my graph of x2+y2=16 translated 2 units right and 1 unit down.

In Exercises 9093, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.

  1. 90. The equation of the circle whose center is at the origin with radius 16 is x2+y2=16.

  2. 91. The graph of (x3)2+(y+5)2=36 is a circle with radius 6 centered at (3, 5).

  3. 92. The graph of (x4)+(y+6)=25 is a circle with radius 5 centered at (4, 6).

  4. 93. The graph of (x3)2+(y+5)2=36 is a circle with radius 6 centered at (3, 5).

  5. 94. Show that the points A(1, 1+d), B(3, 3+d), and C(6, 6+d) are collinear (lie along a straight line) by showing that the distance from A to B plus the distance from B to C equals the distance from A to C.

  6. 95. Prove the midpoint formula by using the following procedure.

    1. Show that the distance between (x1, y1) and (x1+x22, y1+y22) is equal to the distance between (x2, y2) and (x1+x22, y1+y22).

    2. Use the procedure from Exercise 94 and the distances from part (a) to show that the points (x1, y1), (x1+x22, y1+y22), and (x2, y2) are collinear.

  7. 96. Find the area of the donut-shaped region bounded by the graphs of (x2)2+(y+3)2=25 and (x2)2+(y+3)2=36.

  8. 97. A tangent line to a circle is a line that intersects the circle at exactly one point. The tangent line is perpendicular to the radius of the circle at this point of contact. Write an equation in point-slope form for the line tangent to the circle whose equation is x2+y2=25 at the point (3, 4).

Retaining the Concepts

  1. 98. Determine whether the graph of x2y3=2 is symmetric with respect to the y-axis, the x-axis, the origin, more than one of these, or none of these.

    (Section 1.3, Examples 2 and 3)

  2. 99. Determine whether each relation is a function. Give the domain and range for each relation.

    1. {(1, 6), (1, 7), (1, 8)}

    2. {(6, 1), (7, 1), (8, 1)}

    (Section 1.2, Example 2)

  3. 100. Solve: 2x+34x+5=6x2+8x+15.

    (Section P.7, Example 3)

Preview Exercises

Exercises 101103 will help you prepare for the material covered in the next section.

  1. 101. Write an algebraic expression for the fare increase if a $200 plane ticket is increased to x dollars.

  2. 102. Find the perimeter and the area of each rectangle with the given dimensions:

    1. 40 yards by 30 yards

    2. 50 yards by 20 yards.

  3. 103. Solve for h: πr2h=22. Then rewrite 2πr2+2πrh in terms of r.